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JOHN  WILEY  &  SONS 

PUBLISH, 

By  the  same  Author, 

THE  ELEMENTARY  PRINCIPLES  OP  MECHANICS.     By  Prof.  A.  J, 
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CATA.-L.OOTJEJi<     A.lVr>     C;iR<:;TJLA.FlJS     OllA^TIS. 


THE 

ELEMENTARY    PEINCIPLES 


OF 


MECHANICS. 


Vol.  I. 
KINEMATICS. 


BY 

A.  JAY  DU  BOIS,  C.E.,  Ph.D., 

Piofessor  of  Civil  Engineering 

in  the  Sheffield  Scientific  School  of  Yale  University  f 

Author  of  '^Elements  of  Graphical  Statics," 

" The  Strains  in  Fiamed  Structures,'^ 

etc. 


FIRST    EDITION, 

FIRST  THOUSAND. 


JOHIV  S.  PRELL 

CiiAl  &  Mechanical  Engineer. 

SAN  FRANCISCO,  CAL, 

NEW  YORK: 

JOHN  WILEY   &   SONS. 

London:    CHAPMAN  &  HALL,   Limited. 

1899. 


Copyright,  1894, 

BY 

A.  JAY  DU  BOIS. 


BOBBRT   ORUUHOND,   KLKCTHOTTPBR  AND  PRINTER,   NEW  YORK. 


Eiigineeruig 
Library 

Qh 

V.I 


NOTE. 


The  large  type  by  itself  constitutes  an  abridged  course. 
Articles  in  small  type  are  for  advanced  students. 
Articles  containing  applications  of  the  Calculus  are  enclosed 
in  brackets. 


MECHANICS. 


TABLE   OF  CONTENTS. 

Vol.   I. 
KINEMATICS. 


INTRODUCTIOK 

CHAPTER   I. 

MJS.A8UREMEN1. 

PAGKS 

Physical  Science.  Definition  of  Kinematics,  Dynamics,  Statics,  Kinetics, 
Mechanics.  Measurement.  Statement  of  a  Quantity.  Unit.  De- 
rived Unit.  Dimensions  of  a  Derived  Unit.  Meaning  of  Per. 
Homogeneous  Equations.  Unit  of  Time.  Unit  of  Length.  Unit  of 
Mass.  Standard  Units.  Unit  of  Angle.  Curvature.  Tables  of 
Measures 1  -9 


CHAPTER  II. 

POSITION.    TERMS  AND  DEFINITIONS. 

Point  of  Reference.  Position  of  a  Point.  Polar  Co-ordinates.  Car- 
tesian Co-ordinates.  Dimensions  of  Space.  Configuration.  Rigid 
System.  Rest.  Motion.  Path  of  a  Point.  Motion  of  Translation. 
Motion  of  Rotation.  Combined  Translation  and  Rotation.  Material 
Point  or  Particle.     Material  System 10-14 

GENERAL  PRINCIPLES  OF  KINEMATICS. 
CHAPTER  I. 

SPEED. 

Mean  and  Instantaneous  Speed.  Dimensions  of  Unit  of  Speed.  Numeric 
Equations  of  Speed.  Sign  of  Speed.  Speed  a  Scalar  Quantity. 
Homogeneous  Equations.     Examples 15-23 


VI  CONTENTS. 

CHAPTER  II. 

BATE  OF  CHANGE  OF  SPEED. 

PAGES 

Mean  and  Instantaneous  Rate  of  Change  of  Speed.  Dimensions  of 
Unit  of  Rate  of  Change  of  Speed.  Numeric  Equations  of  Rate  of 
Change  of  Speed.  Sign  of  Rate  of  Change  of  Speed.  Kate  of 
Change  of  Speed  a  Scalar  Quantity.  Examples.  Equations  of 
Motion  of  a  Point  under  Different  Rates  of  Change  of  Speed. 
Graphic  Representation  of  Rate  of  Change  of  Speed.    Examples . .     24-33 

CHAPTER  III. 

DISPLACEMENT. 

Line  Representative  of  Displacement.  Relative  Displacement.  Polygon 
and  Triangle  of  Displacements.  Resolution  and  Composition  of 
Displacements.  Analytical  Determination  of  the  Resultant  for 
Concurring  Displacements.     Examples 34-41 

CHAPTER  IV. 

VELOCITY. 

Mean  and  Instantaneous  Velocity.  Unit  of  Velocity.  Uniform  and 
Variable  Velocity.  Velocity  a  Vector  Quantity.  Line  Represen- 
tative of  Velocity.  Polygon  and  Triangle  of  Velocities.  Compo- 
sition and  Resolution  of  Velocities.  Sign  of  Components  of  Ve- 
locity. Analytical  Determination  of  the  Resultant  for  Concurring 
Velocities.     Examples 42-47 

CHAPTER  V. 

ACCELERATION. 

Mean  and  Instantaneous  Acceleration.  Unit  of  Acceleration.  Uniform 
and  Variable  Acceleration.  Acceleration  a  Vector  Quantity.  Line 
Representative  of  Acceleration.  Polygon  and  Triangle  of  Accelera- 
tions. Composition  and  Resolution  of  Accelerations.  Sign  of 
Components  of  Acceleration.  Analytical  Determination  of  the  Re- 
sultant for  Concurring  Accelerations.  Equations  of  Motion  of  a 
Point  under  Different  Accelerations.  The  Hodograph.  Tangential 
and  Normal  Acceleration.     Examples 48-59 


CHAPTER  VL 

MOMENT  OF  DISPLACEMENT,   VELOCITY  OR  ACCELERATION. 

Moment  about  a  Given  Point  or  Axis.  Line  Representative  of  Moment 
of  Displacement,  Velocity  or  Acceleration.  Composition  and  Reso- 
lution of  Moments.  Sign  of  Components  of  Moments.  Algebraic 
Sum  of  Moments  of  Components  equal  to  Moment  of  the  Resultant 
Analytical  Determination  of  Resultant  Velocity  and  Moment  fo'- 
Concurring  Velocities.     Examples 60-70 


VOL.  I. — KINEMATICS.  VU 

CHAPTER   VII. 
ANGULAR    SPEED. 

PAGES 

Angular  Revolution  of  a  Point.  Mean  and  Instantaneous  Angular 
Speed.  Numeric  Equations  of  Angular  Speed.  Sign  of  Angular 
Speed.  Rate  of  Change  of  Angular  Speed.  Numeric  Equations 
of  Rate  of  Change  of  Angular  Speed.  Sign  of  Rate  of  Change 
of  Angular  Speed.  Equations  of  Motion  of  a  Point  under  Different 
Rates  of  Change  of  Angular  Speed.  Angular  Speed  in  Terms  of 
Linear  Speed.  Rate  of  Change  of  Angular  Speed  in  Terms  of 
Tangential  Linear.  Moment  of  Linear  Velocity  in  Terms  of 
Angular  Speed.  Moment  of  Linear  Tangential  in  Terms  of 
Angular  Speed.  Normal  Linear  in  1'erms  of  Angular  Speed. 
Motion  in  a  Circle.  Graphic  Representation  of  Rate  of  Change 
of  Angular  Speed.     Examples 71-80 


CHAPTER  VIII. 

DIFFERENTIAL  EQUATIONS  OF  MOTION  OF  A  POINT. 

Free  Motion  of  a  Point,  Rectangular  Co-ordinates.  Application  of 
Formulas.  Differential  Polar  Equations  for  Motion  of  a  Point  in 
a  Plane.  General  Polar  Equations  for  Motion  of  a  Point  in  a 
Plane,  Acceleration  Central.  Differential  Equations  for  Constrained 
Motion  of  a  Point  in  a  Plane 81-90 


KINEMATICS   OF  A  POINT. 

CHAPTER   I. 
RECTILINEAR  MOTION.     CENTRAL  ACCELERATION. 

Uniform  Acceleration,  Motion  Rectilinear.  Value  of  g.  Formulas  for 
a  Body  Vertically  Projected.  Application  of  Calculus.  Examples. 
Acceleration  Inversely  as  the  Square  of  the  Distance  from  a  Fixed 
Point.     Application  of  Calculus  91-102 

CHAPTER  IL 

RECTILINEAR  SIMPLE  HARMONIC  MOTION.    RESISTING  MEDIUM. 

Simple  Harmonic  Motion.  Simple  Harmonic  Motion  in  a  Straight 
Line.  Amplitude,  Epoch,  Period,  Phase.  Application  of  Calculus. 
Examples.  Body  Falling  iu  a  Resisting  Medium.  Body  Projected 
Upwards  in  a  Resisting  Medium.  Values  of  d,  ^andc.  Examples.  103-116 

CHAPTER  III. 

TRANSLATION  IN  A  CURVED  PATH— PROJECTILES. 

Curved  Path.  Uniform  Acceleration  Inclined  to  Direction  of  Motion. 
Theory  of  Projectiles.  Application  of  Calculus.  Examples. 
Motion  of  Projectile  in  a  Resisting  Medium 117-129 


Viii  CONTENTS. 

CHAPTER  IV 
TRANSLATION  IN  A  CURVED  PATH— HARMONIC  AND  PLANETARY  MOTION. 

PAGES 

Central  Acceleration.  Cases  of  Central  Acceleration.  Cases  of  Har- 
monic Motion.  Resolution  and  Composition  of  Harmonic  Motion. 
Graphic  Representation  of  Harmonic  Motion.  Blackburn's  Pendu- 
lum. Application  of  Calculus.  Planetary  Motion.  Path  for 
Planetary  Motion.  Kepler's  Laws.  Verification  by  Application 
to  the  Moon.  Application  of  Calculus.  Value  of  a' for  Planetary 
Motion.     Examples 130-150 

CHAPTER  V. 

CONSTRAINED  MOTION  OF  A  POINT. 

Motion  on  an  Inclined  Plane.  Uniform  Acceleration.  Examples.  Mo- 
tion in  a  Curved  Path,  Uniform  Acceleration.  Motion  in  a  Circle, 
Uniform  Acceleration.  Examples.  Motion  in  a  Cycloid,  Uniform 
Acceleration.    Application  of  Calculus.   The  Brachistochrone. . . .   150-160 

Miscellaneous  Problems 161-168 


KINEMATICS  OF  A  EIGID  SYSTEM. 

CHAPTER  I. 

RIGID  SYSTEM  WITH  ONE  POINT  FIXED-ROTATION. 

Rotation.  Angular  Displacement  of  a  Rigid  System.  Linear  Dis- 
placement in  Terms  of  Angular.  Line  Representative  of  Angular 
Displacement  of  a  Rigid  System.  Composition  and  Resolution  of 
Angular  Displacements.  Sign  of  Components  of  Angular  Dis- 
placement. Axis  of  Rotation.  Mean  and  Instantaneous  Angular 
Velocity  of  a  Rigid  System.  Instantaneous  Axis  of  Rotation. 
Unit  of  Angular  Velocity.  Unifonn  and  Variable  Angular  Ve- 
locity. Mean  and  Instantaneous  Angular  Acceleration  of  a  Rigid 
System.  Instantaneous  Axis  of  Angular  Acceleration.  Com- 
position and  Resolution  of  Angular  Velocities  and  Accelerations. 
Unit  of  Angular  Acceleration.  Relations  between  Angular  and 
Linear  Velocity  and  Acceleration.  Equations  of  Motion  of  a 
Rotating  Rigid  System  under  Different  Angular  Accelerations. 
Moment  of  Angular  Displacement,  Velocity  or  Acceleration.  Line 
Representatives  of  Angular  Displacement,  Velocity  or  Accelera- 
tion. Condition  for  Rotation  only.  General  Analytical  Deter- 
mination of  Resultant  Angular  Displacement,  Velocity  or  Ac- 
celeration for  any  Number  of  Concurring  Components.  Result- 
ant of  Two  Concurring  Component  Angular  Displacements, 
Velocities  or  Accelerations.     Examples 169-185 

CHAPTER  II. 

TRANSLATION  AND   ROTATION. 

Moment  of  a  Couple.  Composition  and  Resolution  of  Translation 
and   Angular  Displacement.     Displacement  of  a   Rigid   System. 


VOL.  I.  —  KINEMATICS.  IX 


Composition  and  Resolution  of  Translation  and  Angular  Velocity 
or  Acceleration.  Central  Axis.  Screw  Motion.  Centre  of  Paral- 
lel Angular  Velocities.  Rotation  and  Rectilinear  Translation 
Combined.  Combined  Parallel  Rotations,  One  Axis  Fixed. 
Rotation  about  Intersecting  Axes,  One  Axis  Fixed.  Analytical 
Determination  of  Resultant  Angular  Velocity  and  Velocity  of 
Translation  for  a  Rigid  System  with  any  Number  of  Non-concur- 
ring Angular  Velocities.  Parallel  Velocities.  Components  of 
Motion  of  a  Rigid  System.  Composition  and  Resolution  of 
Screws.  Examples.  Relative  Motion  of  a  Body.  Acceleration 
of  Relative  Motion 186-218 


CHAPTER  III. 

Blow  ROTATING  SYSTEM. 

General  Analytical  Relations  for  a  Point  of  a  Rigid  Rotating  System. 

Euler's  Geometric  Equations.     Examples 219-225 

Index .,.., 227-231 


INTRODUCTIOK 


CHAPTEB  L 

MEASUEEMENT. 

Physical  Science. — We  live  in  a  world  of  matter,  space  and  time. 
We  do  not  know  what  these  are  in  themselves  and  we  cannot  ex- 
plain or  define  any  one  of  them  in  terms  of  the  others. 

Thus  we  recognize  matter  in  certain  states  which  we  call  solid, 
liquid  or  gaseous.  We  distinguish  also  different  kinds  of  matter, 
such  as  iron,  wood,  glass,  water,  air,  etc.,  which  we  call  substances. 
We  also  recognize  limited  portions  of  matter  of  definite  shape  and 
volume,  such  as  a  pebble,  a  rain-drop,  a  planet,  etc.,  which  we  call 
material  bodies.     But  what  matter  is  in  itself  we  do  not  know. 

We  also  recognize  matter  as  occupying  space  and  we  note  suc- 
cessive events  as  occupying  time.  But  what  space  and  time  are  in 
themselves  we  do  not  know. 

We  also  recognize  force  as  causing  change  of  motion  of  matter. 
But  what  force  is  in  itself  we  do  not  know. 

Yet  although  we  thus  know  nothing  of  matter,  space,  time  and 
force  in  themselves,  we  can  and  do  investigate  them  in  their  measur- 
able relations,  and  such  investigation  is  the  object  of  all  physical 
science. 

Mechanics — Kinematics  and  Dynamics — Statics  and  Kinetics. — 
That  branch  of  physical  science  which  treats  of  the  measurable 
relations  of  space  alone  is  called  geometry. 

That  which  deals  with  the  measurable  relations  of  space  and 
time  only,  that  is  with  pure  motion,  is  called  kinematics  {Kivrffi  >, 
motion).     To  the  ideas  of  geometry  it  adds  the  idea  of  motion. 

That  which  deals  with  the  measurable  relations  of  space,  time 
and  matter,  involved  in  the  study  of  the  motion  of  material  bodies 
under  the  action  of  force,  is  called  dynamics  {8vvai.ni^  force).  To 
the  ideas  of  kinematics  it  adds  the  idea  of  force. 

We  divide  dynamics  into  two  parts  :  statics,  which  treats  of 
material  bodies  at  rest  under  the  action  of  force,  and  kinetics, 
which  treats  of  material  bodies  in  motion  under  the  action  of 
force.  Statics  is  thus  a  special  case  of  dynamics  which  it  is  con- 
venient to  consider  separately. 

In  the  study  of  machines,  or  of  moving  bodies  generally,  under 
the  action  of  force,  we  have  to  make  use  both  of  the  principles  of 


2  INTRODUCTION".  [CHAP.    I. 

kinematics  and  of  dynamics.  The  term  mechanics  is  therefore  used 
to  include  the  general  principles  of  both  kinematics  and  dynamics, 
while  their  special  application  to  machines  is  called  applied  me- 
chanics or  mechanism.  We  treat  in  this  work  of  mechanics  as  thus 
defined,  or  the  general  principles  of  kinematics  and  dynamics. 
We  have  then  as  the  scheme  of  the  present  work  : 

C  Vol.      I,  Kinematics ; 
Mechanics:  -s  Vol.    II,  Statics;      / -n^^ ««„;«» 
(Vol.  m;  Kinetics;  [  I>y^amics. 

Measurement. — Since  then  we  have  to  do  in  all  that  follows  with 
the  measurable  relations  of  force,  matter,  space  and  time,  the  sub- 
ject of  the  measurement  of  these  quantities  should  first  engage  our 
attention. 

Unit. — In  order  to  measure  any  quantity  whatever,  we  must 
always  compare  its  magnitude  with  the  magnitude  of  another 
quantity  of  the  same  kind.  The  quantity  thus  taken  as  a  standard 
of  comparison  is  called  the  unit  of  measurement. 

Thus  the  unit  of  length  must  itself  be  some  specified  length,  as, 
for  instance,  one  foot,  one  yard,  one  centimeter  or  one  meter.  The 
unit  of  time  must  be  a  specified  time,  as  one  second.  The  unit  of 
mass  must  be  a  specified  mass,  as  one  pound  or  one  gram  or  one 
kilogram. 

The  units  of  mass,  length  and  time  are  called  fundamental  units, 
because  not  derived  from  any  others. 

Statement  of  a  Quantity.  — The  complete  statement  of  a  quantity 
requires,  therefore,  a  statement  of  the  unit  adopted  and  also  a  state- 
ment of  the  result  of  comparison  of  the  magnitude  of  the  quantity 
with  the  magnitude  of  the  unit. 

The  result  of  this  comparison  is  always  a  ratio  between  the 
magnitudes  of  two  quantities  of  the  same  kind  and  is,  therefore, 
always  an  abstract  number. 

This  ratio  or  abstract  number  is  called  the  numeric. 

Thus  we  say  3  feet,  4  seconds,  5  pounds.  In  each  of  these  cases 
we  state  both  the  unit  and  the  numeric,  or  ratio  of  the  magnitude 
of  the  quantity  to  that  of  the  unit.  Thus  3  feet  denotes  a  quantity 
whose  magnitude  is  three  times  the  magnitude  of  one  foot. 

So  for  any  quantity.  In  general,  if  L  stands  for  any  length  and 
[L]  stands  for  the  unit  of  length,  we  have  L  =  l[L],  or  the  length 
equals  I  times  the  unit  of  length.  Here  I  is  the  numeric  and  is  an 
abstract  number. 

Again,  if  Tis  a  certain  interval  of  time,  and  [T]  stands  for  the 
unit  of  time,  we  have  T=  t[T],  or  the  time  equals  t  times  the  unit 
of  time.    Here  t  is  the  numeric  and  is  an  abstract  number. 

So  also  if  ilf  is  a  certain  mass  and  [M]  stands  for  the  unit  of 
mass,  we  have  M=  m[M],  or  the  mass  equals  m  times  the  unit  of 
mass.    Here  m  is  the  numeric  and  is  an  abstract  number. 

Derived  Unit.— A  unit  of  one  kind  which  is  derived  by  reference  . 
to  a  unit  of  another  kind  is  called  a  derived  unit. 

Thus  the  unit  of  area  may  be  taken  as  a  square  whose  side  is  one 
unit  of  length,  or  one  square  foot.  The  unit  of  volume  may  be 
taken  as  a  cube  whose  edge  is  the  unit  of  length,  or  one  cubic  foot. 
The  unit  of  speed  may  be  taken  as  one  unit  of  length  per  unit  of 
tim^,  or  one  foot  per  second. 

Such  units  are  derived  units,  while  the  units  of  mass,  space  and 
time,  not  being  thus  derived  from  any  others,  are  fundamental 
units. 


CHAP.    I.]        MEASUREMENT — HOMOGENEOUS   EQUATIONS.  3 

Dimensions  of  a  Derived  Unit.— A  statement  of  the  mode  m 
which  the  magnitude  of  a  derived  unit  varies  with  the  magnitudes 
of  the  fundamental  units  which  compose  it  is  a  statement  of  the 
dimensions  of  the  derived  unit. 

Thus  let  [A]  denote  the  unit  of  area  and  \L]  the  unit  of  length. 
Then  if  A  —  a[A\  is  the  area  of  a  square  whose  side  is  L  =  l[L], 
where  a  and  I  are  abstract  numbers,  we  shall  have  a[A]  —  riLf. 

Now  we  shall  have  the  numeric  equation  a  —  l'\  or  the  number 
of  units  of  area  equals  the  square  of  the  number  of  units  of  length, 
provided  we  have  [  A]  =  \L]'',  or  the  unit  of  area  equal  to  the  square 
of  the  imit  of  length. 

The  statement  [AJ  =  [Ly  is  a  statement  of  the  dimensions  of  the 
unit  of  area. 

Again,  let  [L]  denote  the  unit  of  length  and  [T]  denote  the  unit 
of  time  and  [V]  denote  the  unit  of  speed.  Then  if  Z,  =  Z[iJ  is  any 
distance  and  T  =  t[T]  is  the  time  occupied  in  describing  that 
distance,  and  the  mean  speed  is  F"=  v[V],  we  have 

We  shall  then  have  the  numeric  equation  v  =  y,  or  the  number 

of  units  of  speed  is  equal  to  the  number  of  units  of  length  passed 
over  divided  by  the  number  of  units  of  time  occupied,  provided  we 
have 

L  y  J  (-jry 

or  the  unit  of  speed  equal  to  one  unit  of  length  per  unit  of  time. 
This  is  a  statement  of  the  dimensions  of  the  unit  of  speed. 

Meaning  of  "Per." — It  will  be  observed  that   the   statement 

[ V]  =  iJ  is  read,  "  the  unit  of  speed  is  equal  to  the  unit  of  length 

per  unit  of  time,"  and  the  word  per  is  indicated  by  the  sign  for 
''divided  by.'' 

Now  we  can  divide  the  numeric  I  by  the  numeric  t,  because  these 
are  abstract  numbers.  But  it  would  be  nonsense  to  speak  of  divid- 
ing length  by  time,  or  a  unit  of  length  by  a  unit  of  time.  We  there- 
fore avoid  such  a  statement  by  the  use  of  the  word  per.  If  then 
we  give  to  the  symbol  of  division  this  new  meaning,  we  can  then 
treat  it  by  the  rules  which  apply  to  the  old  meaning,  and  thus  avoid 
the  invention  of  a  new  symbol  by  using  an  old  one  in  a  new  sense. 

Wlienever,  then,  the  tvord  "  per  "  is  used,  it  can  be  replaced  by  the 
sign  of  division. 

Homogeneous  Equations. — The  symbols  in  all  formulas  or  state- 
ments of  the  relations  of  quantities  always  stand  for  the  numerics 
of  these  quantities,  and  the  units  are  always  understood  though  not 
written. 

Thus  such  an  equation  as  u  =  ^  or  Z  =  rHs  a  numeric  equation, 

and  the  units  are  understood  and  must  always  be  supplied  in  inter- 
preting them.  When  the  units  are  thus  supplied,  all  the  terms  on 
both  sides  of  the  equation  which  are  combined  by  addition  or  sub- 
traction must  always  be  of  the  same  kind,  whatever  the  system  of 
units  adopted.    Such  equations  are  called  homogeneous. 

If  any  numeric  equation  is  not  thus  homogeneous,  it  is  incor- 
rectly stated. 


■4  INTRODUCTIOJSr.  [CHAP,    I. 

It  is  also  evident  that  all  algebraic  combinations  of  such  homo* 
;geneous  equations  must  always  produce  homogeneous  equations. 
If  not,  some  error  must  have  been  made  in  the  algebraic  work. 

Error  can  thus  often  be  detected  in  the  result  of  an  investigation 
without  following  through  its  successive  steps,  by  simply  inserting 
the  omitted  units,  and  no  equation  or  result  should  be  accepted,  or 
even  discussed,  which  does  not  stand  this  test. 

Thus  in  the  equation  I  —  vt,  if  we  supply  the  omitted  units,  we 
iiave 

l[L\  =  v"^  X  t[T]  =  vt[L]. 

The  equation  is  therefore  homogeneous,  since  the  unit  of  length  is 
to  be  understood  in  both  terms. 

Unit  of  Time. — The  unit  of  time  ordinarily  adopted  in  dynamics 
is  the  second  or  some  multiple  of  the  second. 

It  is  the  time  of  vibration  of  an  isochronous  pendulum  which 
T^ibrates  or  beats  86400  times  in  a  mean  solar  day  of  24  hours,  each 
hour  containing  60  minutes  and  each  minute  60  seconds  (24  x  60 
^  60  =  86400). 

The  sidereal  day  contains  86164.09  of  these  mean  solar  seconds. 

Unit  of  Length.  —  The  unit  of  length  ordinarily  adopted  in 
dynamics  is  the  foot  or  the  meter  or  some  multiple  of  these. 

Unit  of  Mass. — The  unit  of  matter  or  mass  ordinarily  adopted  in 
•dynamics  is  the  pound  or  the  kilogram. 

Standard  Unit. — All  units  adopted  are  defined  by  reference  to 
certain  standard  units.  A  standard  unit,  in  general,  should  possess, 
so  far  as  possible,  a  permanent  magnitude  unchanged  by  lapse  of 
time  and  unaffected  by  the  action  of  the  elements  or  by  change  of 
place  or  temperature.  It  should  be  capable  of  exact  duplication 
and  should  admit  of  direct  and  accurate  comparison  with  other 
quantities  of  the  same  kind. 

Standard  Unit  of  Time. — The  standard  unit  of  time  is  the  period 
of  the  earth's  rotation,  or  the  sidereal  day.  This  has  been  proved 
by  Laplace,  from  the  records  of  celestial  phenomena,  not  to  have 
changed  by  so  much  as  one  eight-millionth  part  of  its  length  in  the 
course  of  the  last  two  thousand  years. 

The  length  of  the  solar  day  is  variable,  but  the  mean  solar  day, 
which  is  the  exact  mean  of  all  its  different  lengths,  is  the  period 
already  mentioned,  which  furnishes  the  second  of  time.  It  is 
1.00273791  of  a  sidereal  day. 

The  second  can  therefore  be  defined,  with  reference  to  the  stand- 
ard unit  of  time,  as  the  time  of  one  swing  of  a  pendulum  so  ad- 
justed as  to  make  86400  oscillations  in  1.00273791  of  a  sidereal  day. 

Standard  Units  of  Length. — The  English  standard  unit  of  length 
is  the  length  of  a  standard  bronze  bar,  deposited  in  the  Standards 
Department  of  the  Board  of  Trade  in  London. 

Since  such  a  bar  changes  in  length  with  its  temperature,  the 
length  is  taken  at  the  specified  temperature  of  62°  Fah. 

The  length  of  this  bar  at  this  temperature  is  the  English  stand- 
ard unit  of  length,  and  is  called  the  standard  yard.  Accurate 
copies  of  this  standard  are  distributed  in  various  places,  and  from 
these  all  local  standards  of  length  are  derived.* 

The  foot  is  defined  as  one  third  the  length  of  the  standard  yard 
at  62°  Fah. 

*  The  English  standard  yard  is  1  part  in  17230  shorter  than  the  U.  S.  copy. 


CHAP.   I.]  MEASUREMENT — UNIT   OF   ANGLE.  5 

The  French  standard  of  length  is  the  meter,  and  is  the  length  of 
a  bar  of  platinum  at  the  temperature  of  melting  ice,  or  0°  C.  This 
bar  is  preserved  at  Paris.  Its  length  was  intended  to  be  the  ten- 
millionth  part  of  a  quadrant  of  the  earth's  meridian  through 
Paris. 

The  quadrant  of  the  meridian  through  Paris  is  10001472  stand- 
ard meters,  according  to  Colonel  Clarke's  determinations  of  the 
size  and  figure  of  the  earth,  which  are  at  present  the  most  authori- 
tative, and  thus  the  standard  Paris  meter  is  slightly  less  than  the- 
length  upon  which  it  was  founded.  The  material  bar  is  therefore- 
the  standard,  just  as  is  the  case  with  the  English  standard. 

The  relation  of  the  meter  to  the  meridan  was  intended  as  a 
means  of  reproduction  in  case  of  destruction  of  the  standard,  but 
in  such  case  the  standard  would  probably  be  reproduced  from  the 
best  existing  copies. 

This  was  actually  the  case  with  the  original  English  standard, 
which  was  destroyed  by  fire  in  1834.     It  had  been  originally  de- 

fined  as  having  at  62°  Fah.  a  length  of  -  of  the  length  of  a 

pendulum  vibrating  seconds  in  the  latitude  of  London  at  the  sea- 
level.  But  this  provision  for  its  restoration  was  repealed  and  a 
new  standard  bar  was  constructed  from  authentic  copies  of  the  old 
one. 

The  English  inch,  or  the  36th  part  of  the  length  of  the  standard 
yard,  is  very  nearly  equal  to  the  five-hundred-millionth  part  of  the 

length  of  the  earth's  polar  axis  [-^q^^q) 

The  utUity  of  the  standard,  however,  does  not  depend  upon  any 
such  earth  relations,  the  only  value  of  which  is  for  reproduction  in 
case  of  destruction — a  value  which,  as  we  have  seen,  is  practically 
disregarded. 

The  ultimate  standards  are  therefore  the  actual  bars. 

Standard  Units  of  Mass.^The  English  standard  unit  of  mass  is 
a  piece  of  platinum  deposited  in  the  Office  of  the  Exchequer  at 
London  and  called  the  "Imperial  Standard  Pound  Avoirdupois." 

The  French  standard  unit  of  mass  is  a  piece  of  platinum  pre- 
served at  Paris  and  called  the  kilogram. 

Unit  of  Angle. — There  are  two  units  of  angle  in  use,  the  degree 
and  the  radian. 

The  degree  is  that  angle  subtended  at  the  centre  of  any  circle 

by  an  arc  equal  in  length  to  ^^  part  of  the  circumference  of  that 

circle.  It  is  subdivided  sexagesimally  into  degrees,  (°),  minutes  ('), 
and  seconds  (").  The  seconds  are  subdivided  decimally.  Minutes 
and  seconds  of  time  are  distinguished  by  being  written  mm.,  sec. 

The  radian  i«  that  angle  subtended  at  the  centre  of  any  circle  by 
an  are  equal  in  length  to  the  radius.     It  is  subdivided  decimally. 

If  then  the  length  of  any  arc  is  s[L],  or  s  units  of  length,  and 
the  length  of  the  radius  is  r[L],  or  r  units  of  length,  and  if  the 
angle  subtended  at  the  centre  is  6  radians,  we  have 

„     slL]      s  , 

6  =  -t— ^  =  - ,   or    rB  =s. 

r{L\      r 

The  number  of  radians  in  any  angle  is  then  found  by  dividing  the- 
number  of  units  of  length  in  the  subtending  arc  by  the  number  of 
units  of  length  in  the  radius,  and  this  number  is  independent  of  the 
particular  unit  of  length  adopted,  whether  feet  or  centimeters. 


6  INTKODUCTION.  [CHAP.   I. 

If  the  subtending  arc  is  the  entire  circumference,  the  number  of 
radians  is  —  =  2*.   Hence  27t  radians  correspond  to  360  degrees,  or  1 

radian  corresponds  to   —  = =  57.29578  degrees  —  57°  17'  44".8. 

Any  angle  expressed  in  radians  may  then  be  converted  into 

180° 
degrees  by  multiplying  the  number  of  radians  by =  57.29578 

degrees  =  1  radian. 

Any  angle  expressed  in  degrees  may  be  converted  into  radians 

by  multiplying  the  number  of  degrees  by  -—  =  0.0174533  radians 

180 

=  1  degree. 

(1)  Express  12°  84'  56"  in  terms  of  radians;  and  3  radians  in 
terms  of  degrees. 

Ans.  0.2196  radians  ;  171°  53'  14".424. 

(2)  The  radius  of  a  circle  is  10  feet;  what  is  the  angle  subtended  at 
the  center  by  an  arc  of  3  feet  f 

3 
Ans.   -r  radian,  or  17°  11'  19.  "44. 

(3)  How  much  must  a  rail  30  feet  long  be  bent  in  order  to  fit  into 
a  curve  of  half  a  mile  radius  f 

Ans.  5-^  radian,  or  0°  39'  3". 92. 

00 

(4)  Express  45  degrees  in  terms  of  radians,  and  4.5  radians  in 
terms  of  degrees. 

Ans.  J  radians  =  0.7854  radians  ;  257°  49'  51". 636. 
4 

(5)  The  avgle  subtended  at  the  centre  of  a  circle  by  an  arc  whose 
length  is  1.57  feet  is  15° ;  what  is  the  radius f 

s      157r  157X180      „  „^ 

Ans.   -  =  757, ,    or    r  =  — ^g =  6  ft. 

r      180  15n: 

(6)  What  is  the  sin  —  radians;  cos  —  radians;  cos  —  radians; 

6  o  3 

tan  —  radians  f 
4 

Ans.  0.5;  ^  VS;   0.5;   1. 

(7)  Express  in  degrees  and  in  radians  the  angle  made  by  the 
hands  of  a  clock  at  35  minutes  past  3  o^clock. 

Ans.  102.5  degrees;  1.79  radians. 

Unit  of  Conical  Angle. — Let  the  area  of  any  portion  of  the  sur- 
face of  a  sphere  be  A[A],  or  A  units  of  area,  and  let  the  square 
of  the  radius  be  r'^[A],  or  r''  units  of  area. 

If  lines  are  drawn  from  the  centre  C  of  the  sphere  to  every  point 
of  the  area,  they  form  a  cone,  and  the  angle  subtended  at  the  centre 
C  by  the  area  we  call  a  conical  angle. 

The  conical  angle  subtended  at  the  centre  of  a  sphere  by  a  por- 
tion of  its  surface  whose  area  is  equal  to  the  square  of  its  radius  we 


CHAP.    I.]  MEASUJIEMENT — CURVATURE.  7 

call  a  square  radian.    If  we  denote  the  conical  angle  subtended  by 
the  area  A  by  9  square  radians,  we  have 


A[A] 


=  — ,    or    r^9  =  A. 


r'iA]       r 

The  number  of  square  radians  in  any  conical  angle  is  thu3  found 
by  dividing  the  number  of  units  of  area  in  the  subtending  area  by  the 
number  of  units  of  area  in  the  square  of  the  radium,  and  this  number 
is  independent  of  the  unit  of  area  adopted. 

If  the  subtending  area  is  the  entire  surface  of  the  sphere,  the 

number  of  square  radians  is — 5-  =  4;r.     Hence  the  surface  of   a 

r 

sphere  subtends  a  conical  angle  of  4n:  square  radians. 

[The  terms  solid  angle  and  solid  radian  are  usually  employed  in 
place  of  conical  angle  and  square  radian. as  defined,  but  as  they 
seem  in  no  way  descriptive,  we  have  employed  the  latter  terms  as 
more  expressive.] 

Curvature. — The  direction  of  a  plane  curve  at  any  point  is  that 
of  the  tangent  to  the  curve  at  this  point. 

Thus  the  direction  of  the 
curve  AB  at  the  point  A  is 
that  of  the  tangent  AC. 

The  change  of  direction  be- 
tween any  two  points  of  a 
plane  curve  is  the  angle  be- 
tween the  tangents  at  these 
two  points,  and  is  called  the 
integral  curvature. 

Thus  the  angle  6,  or  change 
of  direction  between  the  tangents  at  A  and  B,  is  the  integral  curva- 
ture for  the  curve  between  A  and  B. 

The  integral  curvature  for  any  portion  of  a  plane  curve  divided  by 
the  length  of  that  portion  is  the  mean  curvature. 

Thus  if  the  length  from  A  to  -B  of  the  curve  is  s[L],  or  s  units  of 

length,  the  mean  curvature  is    ,M--    Since  6  is  given  in  radians,  the 

sill] 

unit  of  curvature  is  one  radian  per  unit  of  length  of  arc.    When  we 
say,  therefore,  that  the  mean  curvature  is  - ,  we  mean  —  radians  per 

unit  of  length  of  arc. 

The  limiting  value  of  the  mean  curvature  when  the  two  points 
are  indefinitely  near  is  called  the  curvature. 

The  curvature,  therefore,  is  the  limiting  rate  of  change  of  direc- 
tion per  unit  of  length  of  arc.  Its  unit  is  one  radian  per  unit  of 
length  of  arc. 

Curvature  of  a  Circle. — If  the  curve  is  a  circle,  the  angle  at  the 
centre  between  the  radii  at  A  and  B  will  be  equal  to  the  angle  9 
between  the  tangents  at  A  and  B. 

We  have  then  9  =  —  radians.     The  naean  curvature   is  then 
r 

6       1 

_  =  —  radians  per  unit  of  length  of  arc. 

s       T 

Since  this  is  independent  of  s,  the  curvature  at  every  point  of  a 
circle  is  constant  and  equal  to  the  mean  curvature  for  any  two 

points,  viz.,  —  radians  per  unit  of  length  of  arc. 
r 


8 


INTRODUCTION. 


[chap,  r. 


Curvature  of  any  Plane  Curve. — For  any  plane  curve  whatever  a 
circle  can  always  be  described  whose  cui-vature  is  the  same  as  that 
of  the  given  curve  at  the  given  point.  This  is  the  circle  of  curvature 
of  the  curve  at  that  point.  Its  radius  is  the  radius  of  curvature  of 
the  curve  at  that  point. 

If  then  p  is  the  radius  of  curvature  of  a  curve  at  any  given 

point,  the  curvature  of  that  point  is  —  radians  per  unit  of  length  of 

arc. 

Since  curvature  then  depends  only  upontheradius  of  curvature, 
the  circle  is  the  only  curve  whose  curvature  is  constant. 

(1)  A  circle  has  a  radius  of  10  feet.     What  is  its  curvature  f 
Ans.     f\f  radian  per  foot  of  arc,  or  5°. 73  per  foot  of  arc. 

(2)  If  the  radius  is  10  yards,  what  is  the  curvature  f 

Ans.  -^Tj  radian  per  yard  of  arc,  or  5°. 73  per  yard  of  arc,  or  -^  radian  per 
foot  of  arc,  or  1°.91  per  foot  of  arc. 

Dimensions  of  Unit  of  Curvature. — If  C  is  the  curvature  and  c 
the  number  of  units  of  curvature,  we  have  by  definition  c[C]  = 

-fJr,  where  [C]  is  the  unit  of  curvature,  and  [0]  is  the  unit  of  angle,  [L] 

s[L] 

the  unit  of  length,  and  6,  s  the  number  of  units  of  angle  and  length. 

We  shall  always  have  c  =  — ,  provided  we  take  [C]  =  f^,  that  is, 

8  [L] 

Erovided  the  unit  of  curvature  is  equal  to  the  unit  of  angle  divided 
y  the  unit  of  length. 
This  is  a  statement  of  the  dimensions  of  the  unit  of  curvature. 
The  unit  of  curvature  is  then   one  unit  of  angle  per  unit  of 
length  of  arc,  as,  for  instance,  one  radian  per  foot  of  arc,  or  one  de- 
gree per  foot  of  arc. 

A  railway  curve  has  a  length  of  one  mile,  the  curvature  is  uniform, 
and  the  integral  curvature  is  30  degrees.  What  is  the  curve,  the 
curvature,  and  the  radius  of  curvature  f 

Ans.     A  circle  ;  0.5336  radian  per  mile  arc  ;  1.9  miles  radius. 

Tables  of  Measures. — We  shall  deal  in  the  course  of  this  work 
with  many  other  derived  units,  which  will  be  explained  as  they 
occur.  It  will  be  useful  to  collect  here  for  convenience  of  reference 
a  number  of  such  units. 


I.    MEASURES    OF   SPACE. 
A.  LENGTH. 

Centimeters 


I    I    I    I 


10 


0 

Table  1. 
1  centimeter  =  0.3937079  inch 
1  Tr,«+o,  _  j  39  37079  inches 

1  meter  -•)  3.2809  feet 

1  kilometer 


2 
Inches 


_  { 0.62137  mile 

"~  (0.535987  nautical  m. 


Table  2. 
1  inch  =  2.539954  centimeters 
1  foot  =  30.479449 
1  yard  =  0.91438347  meter 
1  mile  =  1.60935  kilometers 


CHAP.  I.J     MEASUREMENT — TABLES  OF  MEASURES.  9 

The  following  are  approximate  : 

The  centimeter  is  about  |  inch.     The  meter  is  about  3  ft.  3f  inches. 
The  decimeter  is  about  4  inches.     One  kilometer  is  about  f  of  a  mile. 
Distance  from  pole  to  equator  is  about  10000000  meters. 
Earth's  polar  radius  is  about  500000000  inches. 

B.  AREA. 
Table  3.  Table  4. 

1  sq.  centimeter  =  0.155006  sq.  inch  1  sq.  inch  =  6.45137  centimeters 

1  sq.  meter  =  10.7643  sq.  feet  1  sq.  foot  =  928.997 

1  sq.  hectometer,  I   _  o  ati-ia  a^^^c  1  sq.  vard  =  0.836097  meter 

or  1  hectare        )  "  •^•^'^^*  ^^^^^  1  acre'         =  0.404672  hectare 

1  sq.  kilometer   =  0.38611  sq.  mile  1  sq.  mile  =  2.58989 sq.  kilometers 

C.  VOLUME. 
Table  5.  Table  6. 

1  cubic  centimeter  =  0.0610271  cu.  inch       1  cubic  inch=16.3866cubic  centimeters 
1  liter, or  1  cubic }  =  61 .0271  cu.  inches       1  cubic  foot=28.31o3  liters 

decimeter        f  =1.76172  pints  1  cubic  yard=0. 764513  cubic  meter 

1  cubic  meter         =35.3166  cubic  feet       1  pint  =0.567627  liter 

1  gallon       =4.54102  liters 

II.   MEASURES  OF  MASS. 

Table  7.  Table  8. 

1  centigram  =  0.154323  grain  1  grain  =  0.064799  gram 

1  _  j  15.4323  grains  1  oz.  =  28.3496  grams 

1  gram  _   ^  0.0353739  oz.  1  lb.  or  16  oz.      =  0.453593  kilogram' 

1  kilogram    =  2.20463  lbs.  1  ton  or  2240  lbs.  =  1016.05  kilos 


1  gram  =  mass  of  1  cubic  centimeter  of  pure  water  at  4°  C. 

1  kilogram  =  mass  ot  1  liter  of  pure  water  at  4"  O. 

1  gallon  =  277.274  cubic  inches.  The  gallon  contains  10  lbs.  of  pure  water 
at  62"  F. 

1  cubic  foot  of  water  contains  about  1000  oz.  or  62^  lbs. 

The  pint  contains  20  fluid  oz. 

Acceleration  of  gravity  at  London  =  32.182  feet-per-second  per  second  = 
980.889  centimeters-per-second  per  second.  Average  value  32^  feet-per-second 
per  second  or  980.3  centimeters-per-second  per  second. 

1  dyne  =  force  which  will  give  a  mass  of  1  gram  an  acceleration  of  1  centi- 
meter-per-second  per  second  =  about  j^y  weight  of  gram  =  weight  of  about  1 
milligram. 

1  poundal  =  force  which  will  give  a  mass  of  1  pound  an  acceleration  of  1 
foot-per-second  per  second  =  about  weight  of  |  oz. 


CHAPTER  11. 

POSITION.    TEEMS  AND  DEFINITIONS. 


Point. — A  mathematical  point  has  neither  length,  breadth,  nor 
thickness.  It  is  therefore  without  dimensions  and  indicates  posi- 
tion only. 

Point  of  Reference. — When  we  speak  of  a  point  as  having  position, 
some  other  point  or  points  must  always  be  assumed,  by  reference 
to  which  the  position  is  given.  Such  a  point  is  a  point  of  reference. 
It  is  also  called  a  pole,  or  origin. 

Position  then  is  always  relative.  We  know  nothing  of  "abso- 
lute" position. 

Thus  tlie  position  of  the  point  C  is  known  with 

^  respect  to  A  when  we  know  the  length  of  the  line  AO 

and  the  angle  BA  C  or  the  direction  of  the  line  A  G. 

The  points  A  and  B  are  points  of  reference,  by  means 

of  which  C  is  located. 

Position  of  a  Point.  —The  position  of  a  point 
with  reference  to  other  assumed  points  is  then  known  when  we 
have  sufficient  data  to  locate  it.  These  data  give  rise  to  two 
methods  of  location  : 

1st,  by  polar  co-ordinates. 

2d,  by  Cartesian  co-ordinates,  so  called  because  first  employed  by 
Descartes. 

Plane  Polar  Co-ordinates.— The  data  necessary  for  locating  a 
point  by  polar  co-ordinates,  when  the  point  is  situated  in  a  given 
plane,  consist  of  a  distance  and  an  angle.  If  the  point  is  not  in  a 
known  plane,  of  a  distance  and  two  angles. 

Thus,  if  the  point  P,  in  the  plane  of  this 
page,  is  to  be  located,  we  first  assume  a  line 
OA  in  the  plane,  as  a  line  of  reference.  Then 
the  position  of  P  with  reference  to  O  is  given 
by  the  angle  AOP  and  by  the  distance  OP. 

The  assumed  point  O  is  called  the  pole;  OA 
is  the  line  of  reference;  the  distance  OP  is^ 
called  the  radius  vector,  and  its  magnitude  is 

usually  denoted  by  r  ;  the  angle  AOP  is  the  direction  angle ;  its 
magnitude  is  denoted  by  0,  and  it  is  measured  around  from  OA  to 
the  left. 

The  polar  co-ordinates  for  a  point  in  a  given  plane  are  therefore 
r  and  0,  or  a  distance  and  an  angle.  These  are  plane  polar  co- 
ordinates. 

Space  Polar  Co-ordinates.— If  the  point  P  is  not  in  a  given  plane, 

10 


CHAP. 


II.] 


TERMS  AND    DEFINITIONS. 


11 


we  assume  as  before  a  pole  O,  and  a  reference  line  OA  in  space. 

Through  this  line  we  assume  any 

plane,  as  the  plane  of  this  page, 

OABC,  and  let  OB  be  the  inter- 
section of  this  plane  with  a  plane 

OPB,    perpendicular    to    it     and 

passing  through  OP.    The  location 

of  P  is  then  given  by  the  length 

OP  or    the    radius  vector  r,   the 

angle  AOB  or  (p,  and  the  angle 

BOP  or  Q. 

The  polar  co-ordinates  for  a  point  not  in  a  given  plane  are  there- 
fore r,  (p  and  0,  or  a  distance  and  two  angles.      These  are  spaxie 

polar  co-ordinates. 

If  O  is  a  point  on  the  earth's  surface,  and  the  reference  line  OA 

is  a  north  and  south  line  in  the  plane  of  the  horizon,  the  angles  S 

and  0  would  be  the  astronomical  azimuth  and  altitude  of  the  point 

P. 

Cartesian  Co-ordinates. — Plane.— The  data  necessary  for  locating 

a  point  by  Cartesian  co-ordinates,  if  the  point  is  in  a  known  plane, 

consists  of  two  distances,  parallel  to  two  assumed  lines  of  reference 

in  that  plane,  passing  through  the  point  of  reference,  which  is  called 

the  origin.  The  assumed  lines  of 
reference  are  usually  taken  at  right 
angles. 

Thus  if  the  point  P  is  known  to 
be  in  the  plane  of  this  page,  we 
assume  any  origin  O  and  draw  two 
reference  lines  OX  and  01^  through 
O  in  this  plane  and  at  right  angles. 
These  two  lines  are  called  the  axes 
of  co-ordinates,  the  horizontal  one 
the  axis  of  x,  or  the  x  axis,  the 
other  the  axis  of  y,  or  the  y  axis. 
The  distance  BP  or  OA  is  denoted 

hjx  and  called  the  abscissa  of  the  point  P.    The  distance  APis 

denoted  by  y  and  called  the  ordinate  of  the  point  P. 

The  abscissa  x  is  positive  to  the  ri^ht,  negative  to  the  left  of  the 

origin,  while  the  ordinate  y  is  positive  when  laid  off  above  and 

negative  when  below  the  origin. 

Any  point  in  the  plane  is  thus  located  with  respect  to  O.    If  a 

point  IS  in  the  first  quadrant,  its  co-ordinates  are  +  x,  -f  ^;  if  in  the 

second  quadrant,  —x  and  -i- 

y;  if  in  the  third  quadrant, 

—  X  and  —y;  if  in  the  fourth 

quadrant,  -f  x  and  —  y. 

When  the  point  is   in  a 

known  plane,  the  co-ordi- 
nates are  called  plane  co-or- 
dinates. 

Space     Co-ordinates.  —  If 

the  point  P  is  not  in  a  known 

plane,   we    take    three    axes 

through  the  origin,  all  at  right 

angles  usually.    Two  of  these 

we  may  denote  by  X  and  Y 

as  before ;  the  third,  at  right 

angles  to  the  plane  of  XY,   z 

we  call  the  axis  of  z,  or  the  z  axis. 


1 

•-a; 

r 

p 

+y 

+05 

+2/ 

-y 

o 

3               3 

4 

A 

-y 

3 

1^ 


f 
/ 

/ 

/  1 

^ 

^ 

1 

y       ; 

o 

^' 

X 

/ 

/A 

12  INTRODUCTION.  [CHAP.    11^ 

Thus  the  position  of  the  point  P  is  given  by  the  distance  OA  = 
X,  the  distance  AC  =  z,  and  the  distance  CP  =  y. 
These  are  the  space  coordinates  of  the  point  P. 
The  signs  prefixed  to  the  co-ordinates  indicate  the  quadrant  in 
which  the  point  is  located  as  before.  Thus,  +  x,  +  y  and  ±  z 
denote  a  point  in  the  first  quadrant  either  in  front  of  or  behind  the 
plane  of  XY\  —  x,  +  y  and  ±  z,  a  point  in  the  second  quadrant, 
either  in  front  of  or  behind  the  plane  of  XY;  —  x,  —  y,  ±  z,  and 
+  X,  —  y,  ±  z,  points  in  the  third  and  fourth  quadrants,  either  in 
front  of  or  behind  the  plane  of  XY. 

Direction  Cosines. — If  we  join  the  origin  O  and  the  point  P  by  a 
line,  and  denote  the  angle  of  OP  with  the  x  axis  by  «,  with  the  y 
axis  by  ^,  and  with  the  z  axis  by  r,  we  have  the  relations 

X  =  OP  cos  a,    y  =  OP  cos  /i,    z  =  OP  cos  y. 
These  cosines  are  called  the  direction  cosines  of  OP. 

Since  OP  is  the  diagonal  of  a  parallelogram,  we  have 
OP"  =  a;'  +  2/'  +  2'  =  OP'  (cos"  a  +  cos"  /i  +  cos"  r). 
Hence 

cos"  a  +  cos"  /3  +  cos"  X  =  1 (I)' 

If,  therefore,  any  two  of  these  direction  cosines  are  given,  the 
third  can  always  be  found. 

Since  cos  2a  =  2  cos"  cz  —  1,  cos  2/3  =  2  cos"  /^  —  1,  cos  2y  — 
2  cos"  y  —  1,  yve  have  also 

cos  2a  +  COS  2/3  +  cos  2r  =  —  1 (2) 

Again,  since  cos  (a  +  fi)  =  cos  a  cos  /?  —  sin  «  sin  /S,  cos  {a  —  ft)  = 
cos  a  cos  /3  +  sin  a  sin  ft,  we  have  also 

cos  («  +  ft)  COS  {a  -  /3)  +  cos"  X  =  0 (3> 

Dimensions  of  Space. — A  point  in  a  given  line  is  at  once  located 
by  a  statement  of  the  distance  of  the  point  from  either  end  of  the 
line. 

A  point  in  a  given  plane  is  located  either  by  two  distances  or  by 
a  distance  and  an  angle. 

A  point  in  space  is  located  either  by  three  distances  or  by  a 
distance  and  two  angles. 

Hence  space  is  said  to  have  three  dimensions,  a  plane  surface  to 
have  two  dimensions,  and  a  line  one  dimension. 
A  point  has  no  dimensions  and  indicates 
position' only. 

System.— Any  definite  and  limited  assem- 
_  blage  of  points  is  called  a  system.     Thus  the 
B  assemblage  of  points  represented  by  A,  B,  C,. 
constitutes  a  system. 

Configuration. — The  relative  position  or  arrangement  of  any 
system  or  assemblage  of  points  is  called  the  confignration  of  the 
system. 

A  knowledge  of  the  configuration  of  a  system  at  any  instant, 
requires  a  knowledge  of  the  relative  position  of  every  point  of  the 
system  with  reference  to  every  other  point  at  that  instant. 

Thus  the  configuration  at  any  instant  of  the  system  represented  by  the. 
points  A,  B,  C,  is  known  when  the  position  at  that  c 

instant  of  A  relative  to  B  and  C,  of  B  relative  to  ^ 
and  C,  and  of  C  relative  to  A  and  B,  are  known ; 
that  is,  when  all  the  sides  and  angles  of  the  triangle 
are  known  at  that  instant. 

Rigid   System. — When   the    configuration  ^ 
does  not  change,  the  system  is  a  rigid  system. 


CHAP.    II.]  TERMS   AND    DEFINITIONS.  13 

Thus  when  the  angles  and  sides  in  the  triangle  ABC  remain  un- 
changed, the  system  is  rigid] 

Rest. — When  the  straight  lines  drawn  from  a  point  to  any  as- 
sumed points  of  reference  do  not  change  their  length  or  inclination 
to  each  other,  the  point  is  at  rest  with  reference  to  these  points. 

The  points  of  a  rigid  system  are  therefore  at  rest  relatively  to 
each  other.  The  entire  system  may,  however,  be  in  motion  with 
reference  to  some  external  point. 

Rest,  then,  like  position,  is  relative  only.  We  know  nothing  of 
" absolute"  rest. 

Thus  if  the  lines  CA  and  CB  in  the  preceding  figure  do  not  change  in 
length  and  the  angle  ACB  does  not  change,  the  point  (7  is  at  rest  with  refer- 
ence to  A  and  B. 

The  system  is  then  a  rigid  system,  and  A  is  at  rest  with  reference  to  C  and 
B,  and  B  is  at  rest  with  reference  to  A  and  C. 

The  entire  system  of  ^,  B,  Cmay,  however,  be  in  motion  with  reference  to 
«ome  point  external  to  the  system. 

The  stationary  objects  in  a  room  are  all  at  rest  relatively  to  each  other.  The 
straight  lines  joining  any  three  do  not  change  their  length  or  inclination  to 
each  other. 

But,  as  we  know,  all  these  objects  partake  of  the  motion  of  the  earth,  and 
-are  therefore  not  at  rest  with  reference  to  the  sun. 

Motion. — Motion  is  change  of  position.  A  point  moves  when  the 
straight  lines  joining  it  to  the  points  of  reference  change  either  in 
length  or  inclination  to  each  other. 

Motion,  therefore,  like  rest  and  position,  is  always  relative. 
Such  terms  as  "  absolute  "  position,  "absolute  "  rest,  and  "  absolute" 
motion  have  no  scientific  value.  All  our  knowledge,  both  of  posi- 
tion and  change  of  position,  must  be  essentially  relative. 

"When  a  man  has  acquired  the  habit  of  putting  words  together,  without 
troubling  himself  to  form  the  thoughts  which  ought  to  correspond  to  them,  it 
is  easy  for  him  to  frame  an  antithesis  between  this  relative  knowledge  and  a 
so-called  absolute  knowledge,  and  to  point  out  our  ignorance  of  the  absolute 
position  of  a  point  as  an  instance  of  the  limitation  of  our  faculties.  Any  one, 
however,  who  will  try  to  imagine  the  state  of  a  mind  conscious  of  knowing  the 
absolute  position  of  a  point  will  ever  after  be  content  with  our  relative  knowl- 
■edge."  ("'Matter  and  Motion,"  by  J.  Clerk  Maxwell.  Pott,  Young  &  Co. ,  New 
York,  1876.) 

Path  of  a  Point.  —The  line  joining  the  successive  positions  of  a 
point  during  its  motion  is  called  its  path. 

The  path  thus  described  by  a  point  may  be  either  a  straight  or  a 
<;urved  line,  or  a  combination  of  straight  and  curved  lines.  If  the 
path  is  without  angles  or  abrupt  changes  of  direction,  it  is  con- 
tinuous. 

Motion  of  Translation. —When  a  rigid  system  moves  so  that 
every  straight  line  in  it  joining  every  two  points  remains  always 
parallel  to  itself,  the  system  is  said  to  have  a  motion  of  translation. 

The  paths  of  all  the  points  are  therefore  parallel  at  every  instant 
and  equal  for  any  given  interval  of  time,  and  the  translation  of  the 
system  is  that  of  any  one  of  its  points. 

Motion  of  Rotation. — When  a  rigid  system  moves  so  that  all  its 
points  describe  arcs  of  circles  in  parallel  planes  about  a  common 
straight  line  or  axis  passing  through  the  centers  of  the  circles  and 
perpendicular  to  their  planes,  the  system  is  said  to  rotate  or  have  a 
motion  of  rotation  about  that  axis. 

Since  the  system  is  rigid,  every  point  must  describe  an  equal 
angle  in  the  same  time. 


14  INTRODUCTION.  [CHAP.    II. 

A  point  has  no  dimensions  and  therefore  cannot  have  motion  of 
rotation,  but  only  one  of  translation. 

Combined  Translation  and  Rotation. — A  rigid  system  may  have 
a  motion  of  rotation  and  translation  at  the  same  time.  Thus,  for 
instance,  a  rolling  wheel  has  a  motion  of  rotation  about  an  axis 
through  the  hub  at  right  angles  to  the  plane  of  the  wheel,  while  at 
the  same  time  every  point  of  the  wheel  has  a  motion  of  translation. 

Material  Point  or  Particle. — We  have  just  seen  that  a  point  can- 
not have  motion  of  rotation.  Kotation  is  possible  only  to  systems 
of  points. 

A  material  body  so  small  that  the  distances  between  its  points 
may  be  neglected  is  called  a  particle. 

When  the  body  is  not  small,  whatever  its  magnitude,  if  the  dis- 
tances between  its  various  points  have  no  influence  upon  the  motion 
considered,  we  call  the  body  a  material  point  or  particle  and  may 
represent  it  by  a  point  without  dimensions. 

Thus  if  we  are  investigating  the  motion  of  the  earth  about  the  sun,  so  far 
as  the  motion  of  translation  of  the  earth  is  concerned,  we  may  regard  both  the 
earth  and  sun  as  points.  But  we  cannot  treat  them  as  points  v,  hen  wo  wish  to 
study  their  rotation. 

Material  System.  — A  number  of  material  points  or  particles  con- 
stitute a  material  system.  When  we  confine  our  attention  to  such 
a  system,  all  relations  or  actions  between  one  point  of  such  a  system 
and  another  are  called  internal  relations  or  actions.  Those  between 
the  whole  or  any  part  of  the  system  and  bodies  not  included  in  the 
system  are  called  external  relations  or  actions. 


V^' 


KINEMATICS. 
GENERAL  PRINCIPLES. 


CHAPTER  I. 
SPEED. 


Kinematics. — That  branch  of  science  which  treats  of  the  measur- 
able relations  of  time  and  space  only,  that  is,  of  pure  motion,  is 
called  kinematics.  It  adds  to  the  ideas  of  pure  geometry  the  idea 
of  motion. 

Mean  Speed  of  a  Point, — The  distance  described  by  a  moving 
point  per  unit  of  time  is  called  the  mean  speed  of  the  point.  There- 
fore the  number  of  units  of  distance  described  in  a  given  time, 
divided  by  the  number  of  units  in  that  time,  gives  the  number  of 
units  of  mean  linear  speed.  The  mean  speed  is  then  the  mean 
time-rate  of  motion  in  the  path. 

When  the  mean  speed  varies  with  the  interval  of  time  it  is 
variable.  When  it  has  the  same  magnitude  no  matter  what  the 
intei'val  of  time  it  is  uniform.  A  point  moving  with  uniform  mean 
speed  evidently  describes  equal  distances  in  equal  times. 

Instantaneous  Speed  of  a  Point. — The  limiting  value  of  the  mean 
speed  when  the  interval  of  time  is  indefinitely  small  is  called  the 
instantaneous  speed. 

When  the  instantaneous  speed  at  any  instant  is  equal  to  the 
mean  speed  for  any  interval  of  time  it  is  uniform.  When  the  instan- 
taneous speed  is  variable  the  mean  speed  has  different  values  for 
equal  intervals  of  time. 

The  term  speed  always  signifies  instantaneous  speed  unless 
otherwise  specified. 

Dimensions  of  the  Unit  of  Speed. — Let  us  denote  any  speed  by 
V,  the  unit  of  speed  by  [V]  and  the  number  of  units  of  speed  by  v, 
so  that  V  =v[V].  Then  if  [L]  is  the  unit  of  length  and  s  the  num- 
ber of  units  of  length,  or  the  distance  passed  over  in  the  time  t[T]y 
we  have  by  definition 

^^      t[T] 

o 

We  shall  always  have  the  numeric  equation  v  =  —,  if  we  take 

15 


16  KINEMATICS — GEKERAL   PRINCIPLES.  [CHAP.    I. 

t^]  =  ^»  ^^  *^®  ^"^*  ®^  speed  equals  the  unit  of  length  per  unit  of 

time. 

This  is  the  statement  of  the  dimensions  of  the  unit  of  speed. 
The  unit  of  speed  is  therefore  always  taken  as  one  unit  of  length 
per  unit  of  time,  as,  for  instance,  one  foot  per  second. 

Numeric  Eq^uatious  of  Speed. — If  si  denotes,  then,  the  number  of 
"^  units  in  the  initial  distance  OA,  measured 

along  the  path,  of  a  moving  point  from  a 
fixed  point  O  in  the  path,  taken  as  origin, 
and  s  the  number  of  units  in  the  final 
distance  OB,  measured  along  the  path,  from 
the  same  origin  O,  and  t  the  number  of 
units  in  the  interval  of  time  in  describing 
the  distance  AB  =  s  —  Si,  we  have  for  the  mean  speed  the  numeric 
equation  ^  _  /    ^ 

When  the  interval  of  time  is  indefinitely  small  we  have,  in  the 
notation  of  the  Calculus,  dt  in  place  of  t,  and  ds  in  place  of  s  —  S\. 
The  instantaneous  speed,  or  the  speed,  is  then 

''  =  dt ^'^ 

Sign  of  Speed. — From  (1)  we  see  that  if  OB  or  s  is  numerically 
greater  than  OA  or  s,,  the  value  of  v  will  be  positive,  or  v  equals  a 
plus  (  +  )  quantity.  If,  however,  Si  is  numerically  greater  than  s, 
the  value  of  v  will  equal  a  negative  (— )  quantity.  When,  then,  the 
distance  from  the  origin,  measured  along  the  path,  is  increasing, 
the  value  of  v  is  positive  (+).  When  the  distance  is  decreasing, 
the  value  of  v  is  always  negative  (— ).    Moreover,  if  Si  is  on  the 

S   "4-   Si 

opposite  side  of  the  origin  from  s,  we  have  v  =  — 7 — . 

Equation  (1)  will  therefore  hold  good  generally  if  we  take  dis- 
tances from  the  origin  in  one  direction  as  (+)  and  in  the  other 
direction  as  (— ).  In  such  case,  if  the  value  of  v  comes  out  (  +  )  it 
indicates  motion  in  the  assumed  (+)  direction,  and  if  (— )  it  indi- 
cates motion  in  the  other  direction.  If  t  comes  out  (— )  it  denotes 
time  before  the  start,  if  (-1-)  time  after  the  start  or  beginning  of 
motion. 

Speed  a  Scalar  Quantity.— It  will  be  evident  from  the  preceding 
that  the  sign  of  v  has  no  reference  to  any  special  direction  in 
space.  It  simply  indicates  that  the  distance  along  the  path  from 
the  origin  is  increasing  or  decreasing,  without  reference  to  the 
actual  direction  of  the  path  at  any  instant. 

Speed,  then,  whether  uniform  or  variable,  mean  or  instantane- 
ous, is  independent  of  direction  of  the  path.  A  point  moving  with 
any  given  speed  has  that  speed  no  matter  what  the  shape  of  the 
path. 

Speed,  therefore,  is  a  quantity  which  has  magnitude  and  sign, 
but  is  independent  of  direction.  Such  a  quantity  is  called  a  scalar 
quantity. 

The  student  is  cautioned  here  not  to  confound  speed  with  "velocity," 
which,  as  we  shall  see  hereafter  (page  43),  has  direction  as  well  as  sign  and 
magnitude.     Such  a  directed  quantity  is  called  a  vector  quantity. 


CHAP.    I.]  EXAMPLES— SPEED.  17 

Homogeneous  Equations.— We  have  already  called  attention  in 
the  Introduction  (page  3j  to  the  fact  that  the  units  in  all  numeric 
equations  are  always  understood,  and  when  these  units  are  inserted 
the  equation  must  be  homogeneous,  that  is,  every  term  in  it  must 
stand  for  a  quantity  of  the  same  kind.  When  this  is  not  the  case 
some  error  must  have  been  made  in  the  derivation  of  the  equation, 
and  the  relations  indicated  by  it  are  incorrectly  stated.  By  simply 
inserting  the  units,  then,  in  each  term  of  any  numeric  equation  we 
can  at  once  check  the  result  arrived  at  and  often  discover  without 
further  investigation  if  the  result  is  incorrect. 

Thus  suppose  that  the  result  of  some  investigation  is  expressed 
by 

3s  +  2«  =  lOu. 

Without  reference  to  the  various  steps  by  which  this  result  may 
have  been  reached,  we  can  at  once  say  that  the  result  is  incorrect. 
Thus  if  we  insert  the  vmits,  we  have 

Bs[L] +2tlT-\=  lOv^^, 

and  we  see  at  once  that  the  quantities  in  each  term  are  not  the 
same.    The  equation  is  not  homogeneous.    If,  however,  we  had 

Ss  +  2tv=  lOvt, 

this  equation  is  homogeneous,  because  when  we  insert  the  units 
we  have 

3s[L]  +  2t[T]v^-^-  =  10v^^t[T],  or    Ss[L]  +  2tv[L]  =  10vt[L]. 

Here  all  the  terms  are  quantities  of  the  same  kind,  and  the  equation 
is  homogeneous.  The  relation  expressed  by  it  is  possible,  that  ex- 
pressed by  the  first  is  impossible,  because  we  cannot  add  and  sub- 
tract quantities  of  different  kinds.  It  does  not  follow  that  the 
relation  3s  +  2tv  =  lOvt  is  correct.  It  may  still  have  been  incor- 
rectly deduced.  All  we  can  say  is  it  is  not  on  its  face  absurd,  while 
3s  +  2t  =  lOy  is  manifestly  so. 

The  student  should  make  it  a  rule  to  first  test  in  this  manner 
any  equation  the  truth  of  which  is  suspected,  as  it  may  often  save 
him  the  trouble  of  examining  in  detail  the  entire  investigation  by 
which  it  has  been  deduced.  If,  however,  it  stands  this  test,  then 
the  derivation  must  be  examined  also. 

^  EXAMPLES. 

><(1)  Water  issties  from  an  orifice  having  an  area  of  cross-section 
denoted  by  a,  with  a  speed  of  v.  If  the  discharge  in  cubic  feet  is 
denoted  by  q,  criticise  the  foi^mtda  q  =  av. 

Ans.  Since  a  is  the  area  of  cross-section,  its  unit  must  be  the  unit  of  area, 
as,  for  instance,  one  square  foot.  The  unit  of  v  is  the  unit  of  speed,  or  one  foot 
per  second.  The  unit  oi  q  by  statement  must  be  the  unit  of  volume,  or  one 
cubic  foot.     We  have  then 

.,         V  ft.  cu.  ft. 

q  cu.  ft.  =  a  sq.  ft.  X  z =  ««  :; 

^  ^1  sec.  1  sec. 

The  equation  is  therefore  not  homogeneous.  We  have  cubic  feet  on  one  side, 
equal  to  cubic  feet  per  second  on  the  other.  If,  however,  q  denotes  discharge 
in  cubic  feet  per  second  instead  of  discharge,  then  the  equation  becomes  homo- 
geneous. 


18  KINEMATICS — GENERAL   PRINCIPLES.  [CHAP.   I« 

Even  then  it  does  not  follow  that  the  equation  is  correct  as  compared  with 
fact.  The  actual  discharge  per  second  may  be  less  than  given  hy  g  =  av. 
But  the  equation  as  corrected  is  homogeneous  and  may  be  correct,  whereas- 
before  we  know  it  is  incorrect,  because  it  states  an  impossible  equality  be- 
tween unlike  quantities. 

(2)  A  passenger  sitting  in  a  railroad  car  counts  n  rails  passed 
V        over  in  20  seconds.    If  the  length  of  a  rail  is  29i  feet,  what  is  the 

&peed  in  miles  per  hour  f 

Ans.  n  miles  per  hour.  The  number  of  rails  in  20  seconds  is  the  number 
of  miles  per  hour. 

(3)  If  the  radius  of  the  earth  is  4000  miles,  and  if  it  describes  a 
path  of  600  millions  of  miles  in  365i  dags,  find  (a)  the  speed  of  a 
point  on  the  equator  with  reference  to  a  fixed  point  at  the  equator, 

,  Y  disregarding  the  motion  about  the  sun  ;  (b)  the  speed  in  the  path  with 

/X  reference  to  a  fixed  point  in  the  path,  disregarding  the  motion  of 

rotation. 

Ans.  J^;^535.89  ft.  per  sec;  (6)  19.01  miles  per  sec.  nearly. 

U:^If  the  unit  of  speed  is  taken  at  30  feet  per  second,  and  the  unit 
of  length  at  20  inches,  what  should  be  the  corresponding  unit  of 
time  f 

Ans.  [  F]  =  Y^  ^^  *^®  ^^^*  °^  speed  mast  always  equal  the  unit  of  length 
_  [L]  _  20  in.       r^.,_20in.X  lsec._  1 
[  V]~  30_ft.  •'"'l-^  J  -    30  X  12  in.    ~  18  ^^'^' 
1  sec. 

(5)  If  1  minute  is  the  unit  of  time  adopted,  and  1  decimeter  per 
second  is  the  unit  of  speed,  what  is  the  unit  of  length  f 

Ans.  IL]  =  [V]XIT]  =  ^-^'  X  60  sec.  =  60  decuneters. 


L^€)Tf> 


The  distance  of  a  moving  point  from  a  fixed  point  measured 
in  its  path  is  given  by  s  —  at  +  bf,  where  s  is  the  number  of  feet 
passed  over  in  the  number  of  seconds  t.  (a)  What  is  the  unit  of  a 
and  b  f  (b)  What  is  the  mean  speed  between  the  beginning  of  the 
Qth  and  the  end  of  the  12th  second  from  the  start  f  (c)  What  is  the 
instantaneous  speed  f 

Ans.  (a)  In  order  that  the  equation  may  be  homogeneous,  a  should  be  given 
in  ft.  per  sec.  units  and  b  in  ft.-per-sec.  per  sec.  units,  (b)  For  ^  =  5  sec.  the 
space  passed  over  is  5a  -\-  25b.  For  f  =  12  it  is  12a  -|-  144&.  The  distance 
passed  over  in  the  interval   12  —  5  =  7  sec.  is  7a  -\-  1195.     Hence  the  mean 

speed  is  — ^ —  a-^  176.     (c)  We  have   for  t  =  ti   Si  =  ati  -)-  bti^   and 

hence  s  —  Si  =  a(i  -  «,)  +  Ht^  -  ti^)  or  7— A"  =  a-\-b{t-{-  U).     When  the  in- 

t  —  ti 

terval  of  time  f  —  <,  is  indefinitely  small  the  instantaneous  speed  is  a  +  36^. 

ds 

By  Calculus —=  a -|~  26^ 

J^^j      M  >f?)  The  distance  of  a  moving  point  from  a  given  point  in  its  path 
/u^^'^^i-^-  is  given  by  s  =^  3  +  St,  where  s  is  the  number  of  feet  passed  over  in 
the  number  of  seconds  t.     (a)  What  is  the  unit  for  3  and  8  ?    (b)  What 
is  the  mean  speed  f     (c)   What  is  the  instantaneous  speed  f 

Ans.  (a)  The  number  3  should  stand  for  3  ft.,  the  number  8  for  8  ft.  per 

sec.     (6)  When  t  =  0,  the  initial  distance  is  Si  =  3  ft.     Therefore  — - — =  8  ft. 

■  per  sec.    (c)  Since  this  is  constant,  the  instantaneous  speed  is  the  same.    Or 


CHAP.    I.]  EXAMPLES — SPEED.  19 


S  — 


we  can  write   —  =  8  ft.  per  sec.  whicli  we  see  is  independent  of  the  interval 


^ 


(8)  Suppose  the  distance  is  given  by  s  =  7t  +  8<^  (a)  What  is  the 
mean  speed  and  the  instantaneous  speed  ?  (6)  What  is  the  mean  speed 
between  the  beginning  of  the  10th  and  the  end  of  the  12th  second  f 
(c)  What  is  the  instantaneous  speed  at  the  end  of  the  Gth  second  f 

Ans.  (a) -^  z=7  -^  8{t  -\-  ti)  =  mean   speed.      7  -j-  IQt  =  instantaneous 

t  —  ti 

speed,     (b)  175  ft.  per  sec.     (c)  103  ft.  per  sec. 

A^  (9)  A  train  runs  40  miles  per  hour  for  half  an  hour,  30  miles  an 

^  hour  for  20  minutes,  and  36  miles  an  hour  for  40  minutes.    Find  its 

m,ean  speed. 

Ans.  36  miles  per  hour. 

(10)  A  point  moves  in  a  circle  whose  radius  is  25  feet  and  makes 
6  revolutions  in  15.708  seconds.     What  is  the  mean  speed  f 
Ans.  60  ft.  per  sec. 

^     tA^  (11)  Reduce  (a)  60  feet  per  minute  to  centimeters  per  second;  (b)  1 

Lo^  kilometer  per  hour  to  centimeters  per  second  ;  (c)  36  feet  per  second 

to  yards  per  minute  ;  {d)  10  yards  per  second  to  kilometers  per  hour. 

,  ,      60  ft        60  X  30.479  cm. 

Ans.  (a)    r- — ^ —  = rr^ =  30.479  cm.  per  sec. 

1  mm.  60  sec. 

,^,       1  km.       100000  cm.       „„  „ 

(6)       :r-^r—  =  -^^zr^ =  27.7  cm.  per  sec. 

1  hr.  3600  sec. 

,  ,       36  ft.       12  yds.       „„„    _, 

,^     10  yds.      10  X  0.0091438  km.      „„  „-^„„  , 

id)   —- —  =  — - — J — r —  =  32.9177  km.  per  hour. 

1  sec.  aSTTTf  '^'"• 

(12)  Compare  the  magnitudes  {a)  of  the  foot  per  second  and  the 
mile  per  hour  ;  (b)  of  the  mile  per  hour  and  the  yard  per  minute. 

1  mile 

,  ,    Ihr.         5280  ft.       1  sec.       5280      22     „  ,      ., 

Ans.  (a)  -TT-s—  =  5^7^7^ X  -r-sr  =  557^  =  T^-     Hence  1  mile  per  hour 

1  ft.        3600  sec.        1  ft.       3600      lo 

1  sec. 

22 

is  to  1  ft.  per  sec.  as  --  to  1,  or  as  1.466  to  1. 

15 

1  mile 

,,,  ThrT       1760  yds.       1  min.      1760      „^,      „         .     m 
(6) - — ^  =    „-,     ■ —  X   .     ,    =  -577-  =  ^H.    Hence  1  mile  per  hour 
1  yd.  60  mm.         1  yd.  60 

1  min. 
is  to  1  yd.  per  min.  as  29^  to  1. 

(13)  A  point  describes  50  feet  in  6  minutes  and  another  point  de- 
scribes 50  centimeters  in  6  seconds.    Compare  their  mean  speeds. 

50  ft. 

6  min.       50  X  30.47945  cm.        6  sec.        30.47945      ^  ^^„      „ 

Ans.  — = ^^ X  ^7^ =  — wiT—  =  0.508.     Hence  the 

50  cm.  360  sec.  50  cm.  60 

6  sec. 
mean  speed  in  the  first  case  is  to  that  in  the  second  as  0.508  to  1. 

(14)  A  man  h  feet  in  height  ivalks  along  a  level  street  at  a  uniform 


C<x^' 


fSS^ 


20 


KINEMATICS — GENBEAL   PRINCIPLES.  [CHAP.   I. 


^ 


speed  of  V  miles  per  hour,  in  a  straight  line  from  an  electric  light  I  ft. 
tn  height.    Find  the  mean  speed  of  the  end  of  his  shadow. 
Ans.  I  —  h  :  vt  ::  I  :  X. 


Hence 


I 
l-h 


V  =  speed  required. 


(15)  A  passenger  in  a  railroad  car  moving  ^l-- 
with  uniform  speed  counts  50  telegraph  poles Jfr 
at  equal  intervals  of  100  ft.  passed   in   one 
minute.     What  is  the  mean  speed  of  the  train  f 
Ans.  56.8  miles  per  hour. 

(16)  Three  planets  describe  paths  which  are  to  each  other  as  15,  19, 
and  12,  in  times  which  are  as  7,  3,  and  5.     Find  their  comparative 


Ans.  225,  665,  and  252. 

(17)  Two  bodies  A  and  B  describe  the  same  path  in  the  same  direc- 
tion, with  uniform  speeds  v  and  vf ,  and  at  the  start  the  distance  be- 
tween them  is  a.  Find  the  time  t  when  they  will  be  at  the  distance  b 
in  the  path,  and  the  distance  of  each  from  the  initial  position  of  A 
at  the  end  of  that  time. 

Ans.  Take,  as  the  example  requires,  the  position  of  A,  when  <  =  0,  as  the 
origin,  and  let  distances  in  front  of  this  origin  be  (-|-)  and  behind  it  be  (— ). 
Then  for  the  distance  of  A  from  the  origin  at  the  end  of  any  time  t  we  have 
«  =  vt.  For  the  distance  of  B  from  the  origin  at  the  end  of  the  same  time,  if 
B  is  initially  in  front  of  the  origin  A,  we  have  s'  =  a-{-  v't;  if  B  is  initially 
behind  the  origin  A,  s'  =  —  a-\-v't.  In  general,  then,  «'=  v't  ±  a,  where  the 
(-|-)  sign  is  taken  for  a,  when  B  is  initially  in  front,  and  the  (— )  sign  when  B 
is  initially  behind  the  origin  A.     We  have,  then,  &  —  s  =:  vt  —  v't  '^  a. 

But  by  the  conditions  of  the  problem  s  —  s'  =b.     Hence 


b  =  vt  —  v't  ^  a,     or    t  = 


b  ±  a 


Substituting  this  value  of  t,  we  have 


b  ±  a 
8  =  V ; 


s'  =  ±  a-\-v' 


,b±a       v'h  ±a/o 


•where  the  (-{-)  sign  or  (— )  sign  for  a  is  taken  according  as  B  starts  ahead  of  or 
behind  A. 

(18)  In  the  preceding  example,  suppose  the  bodies  move  in  the 
path  in  opposite  directions. 

Ans.  For  the  distance  of  A  from  the  origin  at  the  end  of  any  time  t,  we 
have,  as  before,  s  =  vt.  If  B  is  initially  in  front  of  A  and  moves  in  the  oppo- 
site direction,  we  have  s^  =  a  —  v't.  If  B  is  initially  beJiind  A,  we  liave  s'  = 
—  a  —  v't.  In  both  cases,  then,  s'  =  —  v't  ±  a,  where,  as  before,  the  (+)  sign 
is  taken  for  a  when  B  is  initially  in  front  of  A,  and  the  (— )  sign  when  B  is 
initially  behind  A.     We  have,  then,  s  —  s'  =  b  =  vt-\-v't  T  a  ;  hence 


t 


b  ±  a 

v-\-v" 


b  ±  a 

v-\-v' ' 


±av  —  v'b 

V  -{-  v' 


(19)  Required  the  time  when  the  two  bodies  are  together. 
Ans.  In  this  case  6  =  0,  and 


±a 

V  ±  v' 


v  ±  v" 


where  the  (+)  sign  or  (— )  sign  is  to  be  taken  for  a  according  as  B  starts  ahead 
or  behind  A,  and  the  {-\-)  sign  or  (— )  sign  for  v'  according  as  the  bodies  move 
in  opposite  or  in  the  same  directions  in  the  path. 


CHAP.    I.]  EXAMPLES— SPEED.  21 

(20)  Give  finally  the  general  solution. 

.         ,       b  ±  a               b  ±  a        ,       ±  av  ±  v'b 
Ans.  t  = -.     s  =  v— ,     8  = — , 

V  ±  V  V  ±  V  V  ±  V 

■where  a  has  the  (+)  or  (— )  sign  according  as  B  is  in  front  or  beliind  A,  and  d' 
has  the  (-(-)  or  (— )  sign  according  as  the  bodies  move  in  opposite  or  in  the 
same  directions  in  the  path. 

(21)  Two  bodies  A  and  B  move  in  the  circumference  of  a  circle 
of  length  c,  with  uniform  speeds  v  and  v',  the  distance  apart  at  the 
beginning  of  the  time  being  a.  Find  the  time  of  the  nth  meeting,  the 
space  described  by  A  and  B,  and  the  interval  between  two  successive 
meetings. 

Ans.  Placing  6  =  0  in  the  value  for  t  in  example  20,  we  have 

±  <i 

ti  = =  time  of  the  first  meeting. 

v  ±  V 

Then  t^  = ,—  =  time  of  t lie  second  meetinsf, 

V  ±v'  ^ 

ti  = —, —  =  time  of  the  third  meeting. 

v  ±  V 

In  general,  tn  =  , =  time  of  the  nth  meeting. 

°  V  ±  V 

±  a4-(n  —  l)c  ,        ■.    .  ■.      A 

Also,      s  =  vtn  =  •» '— ^-7 —  =  space  described  by  A, 

V  ±v' 

±  av'  +  T(n  —  1)C  -v   J  V      r> 

and         sT  a  = ■ — —, =  space  described  by  B. 

V  ±  V 

The  interval  between  two  successive  conjunctions  is 

c 

02  —  ti  ^  fa  —  fa  — 7  • 

V  ±  V 

We  take  (+)  or  (— )  sign  for  a  according  as  B  is  in  front  of  or  behind  A  at 
start,  and  (+)  or  (— )  sign  for  v'  according  as  the  bodies  move  in  opposite  or 
same  directions  in  the  path. 

(22)  When  the  earth  is  in  that  part  of  its  orbit  nearest  to  Jupiter^ 
an  eclipse  of  one  of  Jupiter'^s  satellites  is  seen  16  min.  30  sec.  sooner 
than  when  the  earth  is  most  remote  from  Jupiter.  The  radius  of  the. 
earth's  orbit  being  92390000  miles,  what  is  the  speed  of  light  f 

Ans^^,>86000  miles  per  sec. 

^f!23)  A  train  of  cars  moving  with  a  speed  of  20  miles  an  hour  had 
been  gone  three  hours,  when  a  locomotive  ivas  dispatched  in  pursuit, 
with  a  speed  of  twenty-five  miles  an  hour.  Find  the  time  of  meeting, 
the  speeds  being  maintained  uniform  during  the  time. 

Ans.  t  =  \2  hours. 

(24)  Had  the  trains  in  the  preceding  example  started  together  and 
moved  in  opposite  directions  around  the  earth,  24840  miles,  in  what 
tim£.  would  they  meet  f 

Ans.  23  days.     (See  example  21.) 

(25)  It  is  just  one  o^clock  by  a  clock.  Find  the  time  elapsed  when 
the  minute  and  hour  hands  will  be  together. 

Ans.  5y\  minutes.     (See  example  21.) 

(26)  The  daily  motion  of  Mercury  in  his  orbit  is  4°. 09239;   tJiat  of 


23  KINEMATICS— GENEKAL   PRINCIPLES.  [CHAP.    I. 

FewMS  1". 60216;   that  of  the  earth  0° .985Q3.     What  are  the  intervals 
between  the  epochs  at  which  Mercury  and  Venus  respectively  will  be 
in  the  same  direction  from  the  sun  as  the  earth  ? 
Ana.  115.876  days  and  583.913  days.     (See  example  31.) 

(27)  A  man  caught  in  a  shower  in  which  the  rain  fell  vertically, 
ran  ivith  a  speed  of  12  feet  per  sec.  He  found  that  the  drops  appeared 

\\         to  strike  his  face  at  an  angle  of  10°  with  the  vertical.     What  was 
the  speed  of  the  drops  f 

Ans.  13  tan  80°  =  68  ft,  per  sec. 

(28)  When  the  path  of  the  earth  in  its  orbit  is  perpendicular  to  a 
line  drawn  from  a  star  to  the  earth,  the  direction  of  the  star  appears 
to  make  an  angle  of  20". 445  with  the  perpendicular  to  the  path  of  the 
earth.  The  speed  of  the  earth  being  68180  miles  per  hour,  ivhat  is  the 
speed  of  light  f 

Ans.  191030  miles  per  sec. 

(29)  Compare  the  speeds  of  two  locomotives,  one  of  which  travels 
897f  miles  in  llf  hours  and  the  other  2Q2j%  miles  in  8f  hours. 

yx  Ans.  91  to  81. 

/     /  '    cut         ^^^  -^  ^^^  makes  a  circuit  of  4i  miles  in  one  hour,  stopping  at 
V^     'x/i.o/^*^*'  stations  Jive  minutes  and  two  minutes  respectively,  and  making 
\x^o^j^  o   -((^^Q^^^y  QffiQp  stoppages  of  an  average  duration  of  10  sec.  each ;  find 
AVyzVC        i^^  average  speed, 
ctl^  **^'?*1_  J      Ans.  Jrr44  miles  per  hour. 

uro^'^tj'^'"''^  U-<^)  An  ordinary  train  takes  ten  hours  to  a  certain  trip,  besides 
two  hours  in  all  of  stoppages.  The  express  goes  50  per  cent  faster, 
and  makes  the  trip  in  4  hours  less.  What  time  does  it  lose  in 
stoppages  ? 

Ans.   1  hour  30  min. 

(32)  A  man  rides  a  certain  distance  and  walks  back  in  six  hours. 
He  could  ride  both  ways  in  3i  hours.  How  long  would  it  take  him 
to  ivalk  both  ways  ? 

Ans.  8^  hours. 

(33)  The  speed  of  the  periphery  of  a  wheel  12  feet  in  diameter  is 
6  feet  per  sec;  find  the  revolutions  per  minute. 

Y y  Anar^.S  rev.  per  min. 

^    ^  <  v  ^         1^34)  A  person  inquiring  the  time  of  day  is  told  that  it  is  between 
^^  ^  '      ^  \-    5  and  6  o^clock,  and  that  the  hour  and  minute  hands  are  together. 

Find  the  time. 
'/  '  '^v-*'   V/^      Ans.  5  hr.  37  min.  16^*^  sec. 

(35)  Find  the  number  of  revolutions  per  mile  made  by  a  wheel 
4i  feet  diameter. 

Ans.  373  rev.  per  mile. 

hi       (36)  How  soon  after  8  o'clock  are  the  hour  and  minute  hands 
'  directly  opposite  f 
Ans.  10^^  min. 

(37)  Two  men  walk  opposite  ivays  round  a  circular  course.  ^  They 
.ji     meet  for  the  first  time  at  the  north  point,  the  sixth  time  at  tJie  east 
'      point.     Where  will  they  meet  for  the  sixteenth  time,  and  what  are 
the  relative  speeds  f 

Ans.  At  the  w^est  point.     Ratio  of  the  speeds,  19  to  1,  or  3  to  1. 


'CHAP.    I.]  EXAMPLES — SPEED.  23 

(38)  A  courier  starts  from  a  given  point  with  a  speed  of  b  miles 
in  a  hours.  After  n  hours  a  second  courier,  travelling  at  the  rate 
of  d  miles  in  c  hours,  sets  out  from  a  point  q  miles  ahead  or  behind 
the  first  point,  and  travels  over  the  same  route.  In  what  time  will 
the  second  courier  overtake  the  first  f    (See  example  19.) 

Ans.  — jC  hours,  where  the  (— )  sign  is  taken  if  q  is  ahead  and  the  (+) 

(X/Q/  —  CO  * 

sign  if  q  is  behind. 

(39)  A  hollow  ball  A  floating  upon  the  surface  of  a  stream  is 
observed  to  have  a  speed  of  Vo.  If  A  is  united  by  a  thin  wire  to 
another  ball  B  which  sinks  in  water,  the  speed  of  A  is  now  observed 
to  be  vi.     Find  the  speed  Vi  of  the  water  at  the  depth  of  B. 

Ans.  We  have  the  combined  speed  v  =  -^-P— \  hence  »i  =  2»  —  ■Oo- 


:fo  'At.t    Kt^^f--' 


CHAPTEE  II. 

RATE  OF  CHANGE  OF  SPEED. 

EQUATIONS  OP  MOTION  UNDER  DIFFERENT  RATES  OF  CHANGE  OF  SPEED 
GRAPHIC   REPRESENTATION   OF  RATE   OF  CHANGE  OF   SPEED. 

Change  of  Speed.— When  the  speed  of  a  point  is  variable,  th© 
difference  between  the  final  and  initial  instantaneous  speeds  for 
any  interval  of  time  is  the  total  or  integral  change  of  speed  for 
that  time. 

Mean  Rate  of  Change  of  Speed.— The  integral  change  of  speed 
per  unit  of  time  is  the  mean  rate  of  change  of  speed.  Therefore  the 
number  of  units  in  the  integral  change  of  speed  for  any  interval  of 
time,  divided  by  the  number  of  units  in  that  time,  gives  the  number 
of  units  in  the  mean  rate  of  change  of  speed. 

When  the  mean  rate  of  change  of  speed  varies  with  the  interval 
of  time,  it  is  variable.  When  it  has  the  same  magnitude,  no  matter 
what  the  interval  of  time,  it  is  uniform. 

Instantaneous  Rate  of  Change  of  Speed. — The  limiting  value  of 
the  mean  rate  of  change  of  speed,  when  the  interval  of  time  is  in- 
definitely small,  is  the  instantaneous  rate  of  change  of  speed. 

Rate  of  change  of  speed  should  always  be  understood  as  mean- 
ing the  instantaneous  rate  of  change  unless  otherwise  specified. 

Rate  of  change  of  speed  may  be  zero,  uniform  or  variable. 
When  it  is  zero,  the  speed  is  uniform  and  is  the  same  as  the  mean 
speed  for  any  interval  of  time. 

When  it  is  uniform,  the  rate  of  change  of  speed  is  the  same  as 
the  mean  rate  of  change  for  any  interval  of  time. 

When  it  is  variable,  the  mean  rate  of  change  has  different  values 
for  equal  intervals  of  time. 

Dimensions  of  Unit  of  Rate  of  Change  of  Speed.— Let  us  denote 
the  rate  of  change  of  speed  by  A  and  its  unit  by  \A  J  and  the  number 
of  units  by  a,  so  that  A  =  a[A\  Then  if  v^\V]  and  u[F]  are  the 
initial  and  final  instantaneous  speeds  and  t[T]  the  corresponding 
interval  of  time,  we  have  the  integral  change  of  speed  {v  —  Ui)[F], 
and  by  definition  the  mean  rate  of  change  of  speed  is 

a[A]-  ^j        . 

We  shall  always  have  the  numeric  equation 

V  —  Vi 


t 
if  we  take 


[A]  =  [^  =  [^i(pagel6). 


CHAP.  II.]  RATE   OF   CHANGE   OF   SPEED.  25 

The  unit  of  rate  of  change  of  speed  is  then  one  unit  of  speed  per 
unit  of  time,  as  for  instance  one  foot-per-sec.  per  sec. 

Numeric  Equations  of  Rate  of  Change  of  Speed.— We  have  then 
the  numeric  equation  for  the  mean  rate  of  change  of  speed 


a  =  —^, (1) 

where  v  and  th  are  the  number  of  units  in  the  final  and  initial 
instantaneous  speeds  during  the  interval  of  time  t  units. 

For  the  instantaneous  rate  of  change  of  speed  we  have  the 
limiting  value  in  the  notation  of  the  Calculus 

dv     d^s 
''  =  dt=dt^ ^2> 

Sign  of  Rate  of  Change  of  Speed.— From  (1)  we  see  that  if  v  is 
numerically  greater  than  Vi  the  value  of  a  will  be  positive  (  +  ),  and 
if  tJi  is  numerically  greater  than  v  the  motion  is  retarded  and  the 
value  of  a  is  negative  (— ).  The  value  of  a  then  is  positive  (+)  when 
the  speed  increases,  and  negative  (— )  when  the  speed  decreases 
during  the  interval  of  time  t. 

Rate  of  Change  of  Speed  a  Scalar  Quantity. — It  is  evident  that 
the  sign  of  a  has  no  reference  to  any  special  direction,  but  simply 
shows  whether  the  speed  is  increasing  or  decreasing  numerically. 
Rate  of  change  of  speed  is  therefore,  like  speed  (page  16),  a  scalar 
quantity. 

The  student  is  cautioned  here  not  to  confound  rate  of  change  of  speed  with 
"acceleration,"  which,  as  we  shall  see  hereafter  (page  49),  has  direction  as  well 
as  magnitude  and  sign,  and  is  therefore  a  vector  quantity. 


WtlT^ 


EXAMPLES. 


[)  A  paint  moves  at  a  given  instant  with  a  speed  of  12  ft.  per 

sec.,  and  at  the  end  of  3  sec.  after,  with  a  speed  of  16  ft.  per  sec. 

What  is  the  change  of  speed,  and  the  mean  rate  of  change  of  speed  f 

Ans.  +4  ft.  per  sec.  ;  -]- 1  ft.-per-sec.  per  sec.    The  (+)  sign  indicates  that 

the  speed  increases. 

(2)  A  train  starts  from,  rest  and  in  8  m,in.  attains  a  speed  of  30 
miles  an  hour.     What  is  the  uniform  rate  of  change  of  speed  f 

Ans.  -f"  225  miles-per-hour  per  hour. 

(3)  Compare  the  foot-per-sec.  per  sec.  and  the  yard-per-min.  per 
min. 

Ans,  1  ft.-per-sec.  per  sec.  is  equivalent  to  1300  yards-per-min.  per  min. 

v(4)  Reduce  400  ft.-per-min.  per  min.  to  kilometers-per-hour  per 
hour,  to  feet-per-sec.  per  sec,  and  to  centimeter s-per-sec.  per  sec. 

Ans.  438.9  kil.-per-hour  per  hour  ;  J  f t. -per-sec.  per  sec.  ;  3.38  cen. -per  sec. 
per  sec. 

(5)  Reduce  1  mile-per-min.  per  min.  t9  centim^eters-per-sec  per  sec. 

Ans.  44.7. 

L^:(ff^~lve  take  32.2  ft.-per-sec.  per  sec.  as  the  unit  of  rate  of 
change  of  speed,  and  5  yards  as  the  unit  of  length,  what  should  he 
the  unit  of  time  f 


Ans.  [T]  =  l/g]  =  1/3^  1  sec*  =  0.68  sec. 


^ 


36>  KINEMATICS — GENERAL  PRINCIPLES.  [CHAP.  II. 

*  (7)  In  15  sec.  the  speed  of  a  point  changes  from  400  to  100  ft.  per 
sec.    Find  the  uniform  mean  rate  of  change  of  speed. 

Ans.  —  20  ft.-per-sec.  per  sec.  The  minus  sign  shows  that  the  speed 
decreases. 

(8)  The  distance  measured  in  the  path  of  a  moving  point  from  a 
fixed  point  is  given  by  s  =  at  +  bt'\  where  s  is  the  number  of  feet 
passed  over  in  the  number  of  seconds  t.  What  is  the  mean  speed, 
instantaneous  speed,  mean  rate  of  change  of  speed,  instantaneous 
rate  of  cliange  of  speed  f 

Ans.    Mean  speed, =  a -\-b{t  -\- ti).     Instantaneous  speed,   —=  a -|- 

t  —  fci  (it 

2bt.     Mean  rate  of  change  of  speed, =  2b.     Instantaneous,  the  same. 

t  —  Ti 

y  (9)  If  the  distance  is  given  by  s  —  2t  +  Sf  +  4t\  what  is  the  mean 
speed  between  the  origin  and  final  position  at  the  end  of  5  sec.  f 
What  is  the  instantaneous  speed  at  the  origin  and  at  the  end  of  5 
sec.  f  What  is  the  mean  rate  of  change  of  speed  between  the  origin 
and  at  the  end  of  5  sec.  f  What  is  the  instantaneous  rate  of  change 
of  speed  at  the  origin  and  at  the  end  of  5  sec.  f 

Ans.   Mean  speed,  *^'  =  3  +  3(«  +  ti)  +  Mt^  +  tt,  +  ^i'^.    If  we  let  «,  = 

0  when  t^  =  0,  the  time  t  counts  from  the  origin,  and  we  have  for  t  =  5  mean 

speed  =  117  ft.  per  sec. 

ds 
Instantaneous  speed,  v  =  ~  =^  2^&t-\-  12i*.     For  t  =  0  this  gives  2  ft.  per 

sec.     For  t  =  5,  332  ft.  per  sec. 

1)  —  7} 

Mean  rate  of  change  of  speed, -'  =  6  -}-  12{t  -f  ti).     For  ^,  =  0  and  t  = 

5,  this  gives  66  ft.-per-sec  per  sec. 

Inst9,ntaneous  rate  of  change  of  speed,  —-  =  Q-\-  2it.  For  ^  =  0  this  gives 
6  ft.-per-sec.  per  sec.     For  t  =  5,  126  ft.-per-sec.  per  sec. 

(10)  If  a  point  traverses  in  t  units  of  time  a  distance  measured 

in  the  path  of  s  units  of  length,  given  by  s  =  -t+  bt',  where  a  and  b 

are  constants,  what  is  the  mean  speed,  the  instantaneous  speed,  the 
mean  rate  of  change  of  speed,  the  instantaneous  rate  of  change  of 
speed  ? 

Ans.  Mean  speed, -'  =  —  —  -\-h{t  -\-  ti). 

Instantaneous  speed,  v  =  —  —  — --{-  2bt. 

Mean  rate  of  change, j-  =  — — bj— ^  +  2o. 

Instantaneous  rate  of  change,  -—  =  -—  -f  26. 

at       f 

(11)  A  steamer  leaves  Liverpool  for  New  York,  and  a  vessel  leaves 
New  YorJc  for  Liverpool  at  the  same  time  ;  they  meet,  and  when  the 
steamer  reaches  New  York,  the  vessel  has  as  far  to  go  as  the  steamer 
had  when  they  met.  If  the  distance  is  3000  miles,  how  far  out  from 
Zdverpool  did  they  meet  f 

Ans.  1854  miles.  ~   ' '• 


OHAP.  II.]  KATE   OF   CHANGE   OF   SPEED.  27 

>  (12)  A  tvalks  at  the  speed  of  3|  miles  per  hour  and  starts  18 

minutes  before  B.    At  what  speed  must  B  walk  to  overtake  A  at  the 
ninth  milestone  f 

Ans.  4.29  miles  per  hour. 

(13)  A  tourist  left  behind  by  his  companions,  wishes  to  rejoin 
them  on  the  following  day.  He  knows  they  are  5  miles  ahead,  ivill 
start  in  the  morning  at  8  o^ clock,  and  will  walk  at  the  rate  of  3i  miles 
an  hour.  When  must  he  start  in  order  to  overtake  them  at  1  o^clock 
p.m.,  walking  at  the  rate  of  4  miles  an  hour,  and  resting  once  for 
half  an  hour  on  the  road  ? 

Ans.  7  h.  12  m.  a.m. 

(14)  A  man  walking  4  miles  an  hour  meets  20  street  cars  in  an 
hour  and  is  overtaken  by  4.  What  is  the  average  speed  of  the  cars, 
and  ivhat  is  the  average  distance  between  successive  cars  f 

Ans.  6  miles  per  hour  ;  ^  mile. 

(15)  J.  starts  from  a  raihvay  station,  loalking  5  m,iles  an  hour  ; 
at  the  end  of  an  hour  B  starts,  ivalking  4  Tniles  an  hour.  At  the  end 
of  another  hour  a  train  starts  and  passes  A  25  minutes  after  it 
passes  B.    Find  the  speed  of  the  train. 

Ans.  20  miles  an  liour. 

^ffo)  A  passenger-train  going  41  rniles  an  hour  and  431  feet  long 
overtakes  a  freight  on  a  parallel  line.     The  freight-train  is  713  feet 
long  and  is  going  28  miles  an  hour.    Hoio  long  does  it  take  the  pas- 
senger-train to  pass  ? 
Ans.  1  minute. 

(17)  In  a  mile  race  A  gives  B  50  yards.  B  passes  the  line  5 
TThinutes  after  the  start.  A  passes  it  5  seconds  later.  Which  would 
win  in  an  even  race,  and  by  what  distance  f 

Ans.  A  by  21^  yards. 

Equations  of  Motion  of  a  Point  under  DiflFerent  Rates  of  Change 
of  Speed. — The  rate  of  change  of  speed  may  be  zero,  uniform  or 
variable.    When  variable,  it  may  vary  according  to  any  law. 

(a)  Rate  of  Change  of  Speed  Zero. — When  the  rate  of  change  of 
speed  is  zero,  the  speed  is  uniform,  and  the  instantaneous  speed  at 
any  instant  is  equal  to  the  mean  speed  for  any  interval  of  time. 

In  this  case,  if  Si  is  the  number  of  units  m  the  initial  distance, 
measured  along  the  path  from  the  origin,  and  s  the  number  of  units 
in  the  final  distance,  we  have 

s  —  Si 

v= — 7 — ,    or    vt  =  s—8i (1) 

This  equation  will  be  general  if  we  take  distances  from  the  origin 
in  one  direction  as  (  +  )  and  in  the  other  direction  as  (— ).  In  such 
case,  if  the  value  of  v  comes  out  ( +)  it  denotes  motion  in  the  assumed 
(  +)  direction;  if  (— ),  it  denotes  motion  in  the  other  direction.  If  t 
comes  out  (  +  )  it  denotes  time  after,  if  (— )  time  before  the  start,  or 
beginning  of  motion. 

(6)  Rate  of  Change  of  Speed  Uniform. — When  the  rate  of  change 
of  speed  is  uniform,  the  instantaneous  rate  of  change  of  speed  at 
any  instant  is  equal  to  the  mean  rate  of  change  of  speed  for  any 
interval  of  time. 

In  this  case,  if  v  and  Vi  are  the  number  of  units  in  the  initial  and 


/ 


28  KINEMATICS — GENERAL    PRINCIPLES.  [CHAP.    II.. 

final  instantaneous  speeds  for  any  interval  of  time  t  units,  we  have 
for  the  uniform  rate  of  change  of  speed 

a  =  — ^ — (2) 

The  value  of  a  is  (+)  when  the  speed  increases  and  (— )  when  it  de- 
creases during  the  interval  of  time  t. 
From  equation  (2)  we  have 

V  =  Vi  +  at (3> 

The  average  speed  during  the  interval  t  is 

V+Vi  1 

mean  speed  =  — ^ —  =  vi  +  ^at (4> 

The  distance  (s  —  Si)  betiveen  the  initial  and  final  positions,  de- 
scribed in  the  time  t,  is  the  mean  speed  multiplied  by  the  time,  or 

s  —  s,  =  — 2 — t  =  v,t  +  -^aP (5> 

Inserting  the  value  of  t  from  (2)  we  have 

v"  —  v," 

'-''  =  ^r- («> 

Hence  v'  =  Vi'  +  2a(s  —  Si) (7) 

In  applying  these  formulas,  a  is  positive  (+),  when  the  speed  in- 
creases, and  negative  (— ),  when  the  speed  decreases,  without  regard 
to  direction  of  motion. 

If  distances  s,  Si ,  from  the  origin,  in  one  direction  are  taken  as 
positive  (+),  distances  in  the  opposite  direction  are  negative  (— ). 

Speeds  v,  v^  are  positive  (  +  )  when  motion  is  in  the  assumed 

fositive  direction,  and  negative  (— )  when  in  the  other  direction, 
f  ^  is  minus  (— ),  it  denotes  time  before  the  beginning  of  motion;  if 
plus  ( -I- ),  time  after. 

By  means  of  these  equations,  if  we  have  given  the  initial  position 
of  a  point  moving  in  any  path,  its  initial  speed  and  uniform  change 
of  speed,  we  can  determine  its  final  position  and  speed  and  the  dis- 
tance described  in  any  given  interval  of  time. 

[(c)  Kate  of  Change  of  Speed  Variable.] — If  the  rate  of  change  of  speed  is 
variable,  we  have  from  (1),  in  Calculus  notation,  for  the  instantaneous  speed, 

^  =  d?= <^> 

and  from  (2), 

dv       <Ps  .„, 

''=di=dl^'' ^^> 

and  from  (8), 

s  -  s,  =  /  vdt (10) 

The  preceding  equations  for  constant  rate  of  change  of  speed  can  be  directly 
deduced  from  these  three  general  equations. 

Thus  if  we  suppose  a  constant,  we  have,  by  integrating  (9),  v  =  at  -^  C,  where 
C  is  the  constant  of  integration.     When  ^  =  0,  we  have  v  =  v^,  and  hence  C  = 


^.-t 


CHAP.  II,] 


RATE   OF   CHANGE   OF   SPEED. 


29 


d    N 


«, .  Hence  v  =  r^  -\-  at,  whicli  is  equation  (3).  Inserting  this  value  of  v  in  (8) 
and  integrating,  we  have  s  —  Vit  -}-  iat'^  -\-  U.  But  when  <  =  0  we  have  «  =  s, , 
hence  C  =  «,,  and,  therefore,  s  —  «,  =  Vit  -{■  iaP.     This  is  equation  (5). 

In  any  case,  if  the  law  of  variation  of  a  is  given,  we  can  find  the  relation 
between  v,  s  and  t. 

Graphic  Representation  of  Rate  of  Change  of  Speed. — If  we  rep- 
resent intervals  of  time  by  distances  laid  off  horizontally  along  the 
axis  of  X,  and  the  corresponding  speeds  by  ordinates  parallel  to  the 
axis  of  y,  we  shall  have  in  general  a  curve  for  which  the  change  of 
y  with  X  will  show  the  law  of  change  of  speed  with  the  time. 

(a)  Rate  of  Change  of  Speed  Zero. — Lay  off  from  A  along  AB 
equal  distances,  so  that  the  distances  from  ^  to  1,  1  to  2,  2  to  3,  etc., 
are  all  equal  and  represent  each  one  second  of  time,  and  let  AB 
represent  the  entii-e  time  f.  y 

Then  at  A,  1,  2,  3,  and  B 
erect  the  perpendiculars  AM,  lb, 
2c,  3d,  BN,  and  let  the  length  of 
each  represent  the  speed  at  the 
corresponding  instant. 

Since  there  is  no  change  of 
speed,  these  perpendiculars  will 
all  be  of  equal  length,  we  shall  "        '''■       •*"•      '^■ 

have  AM  =lb  =  2c^3d  =  BN  ^  t-  " 

=  V,  and  the  speed  at  any  interval  of  time  will  be  given  by  the 
ordinate  at  that  instant  to  the  line  MN  parallel  to  AB. 

The  space  described  in  any  time  is  given  by  s—Si  =  vt.    This  is 

evidently  given  by, the  area,  AMNB  in  the  diagram.    Therefore,  the 

area  corresponding  to  any  time  gives  the  space  described  in  that  time. 

(6)  Rate  of  Change  of  Speed  Constant. — Lay  off  as  before  the  time 

along  AB,  and  at  A,  1,  2,  3,  B, 
the  corresponding  speeds,  so 
that  AM  is  the  initial  speed  r, 
and  BN  the  final  speed  v.  Draw 
MC,  be',  cd',  parallel  to  AB. 

Then  bb'  will  be  the  change 
of  speed  in  the  first  sec,  cc'  the 
change  of  speed  in  the  next  sec, 
x  and  so  on.  Since  these  are  to 
be  constant,  NM  is  a  straight 
line,  the  ordinate  to  which  at 


cc 


dd' 
sec      1  sec. 


any  instant  will  give  the  speed  at  that  instant. 
The  rate  of  change  of  speed  is  then =  - 

But =  — ^'  =  a.  Hence  the  rate  of  change  of  speed  is  the  tan- 
gent of  the  angle  NMC  which  the  line  MN  makes  with  the  horizontal. 
Hence  a  =  —    or   NC  =  at. 

The  distance  described  in  the  time  t  is  from  equation  (5)  given 

V  -i-  "V 

by  — K-^f-     But  this  is  the  area  of  AMNB.    Therefore,  the  area 

corresponding  to  any  time  gives  the  space  described  in  that  time. 
We  have  then  directly  from  the  figure,  since  NC  =  at, 


(ft' 


^-^t  =  vd  +  \nC  xt  =  va  +  lat\ 


30 


KINEMATICS — GENERAL   PRINCIPLES. 


CHAP.    II, 


If  V  is  greater  than  v,  a  will  be  negative,  and  the  line  MN  is  in- 
clined Delow  the  horizontal  MC. 

[(c)  Kate  of  Change  of  Speed  Variable.]— If  the  rate  of  change  of  speed  is 
not  constant,  we  shall  have  in  general  a  curve  MNn.  The  tangent  to  this 
curve  at  any  point  N  makes  an  angle  with 

the  axis  of  X,  whose  tangent  is  -7-  =  a, 

equation  (9),  or  the  rate  of  change  of  speed. 
The  elementary  area  BNiib  =  vdt  =  ds, 
equation  (8),  and  the  total  area  AMNB  = 


£ 


vdt  =  s— »,,  equation  (10). 
t-o 

dv  d's 

When  -—  =  0,   or    -— -  =  0, 

dt         '  dt^ 


^ 


or    a  =  0, 


B    b 


the  tangent  to  the  curve  is  horizontal  at  the  corresponding  point,  and  we  have 
the  speed  at  that  point  a  maximum  or  minimum,  according  as  the  curve  is  con- 
cave or  convex  to  the  axis  of  X. 


EXAMPLES. 

[)  The  speed  of  a  point  changes  from  50  to  30  ft.  per  sec.  in  pass- 
ing over  80  ft.  Find  the  constant  rate  of  change  of  speed  and  the 
time  of  motion. 

e*_B,,       900-2500  ._   ,^  rpi,        • 

=  — " — — —  =.  —  10  ft.-per-sec.  per  sec.      The  minus 


Ans,  a  = 


2(«  -  «j)  2  X  80 


sign  indicates  decreasing  speed,     t  — 


v^  _  30  -  50 
~  ~    -10 


=  2  sec. 


(2)  Draw  a  figure  representing  the  motion  in  the  preceding  exam- 
ple, and  deduce  the  results  directly  from  it. 

Ans.  Average  speed  = x —  =  40  ft.  per  sec.     Hence 


^-30  40«  =  80,  or  t 
sec.  per  sec. 


2  sec.      Also  a  = 


30  -  50 
2 


=  -  10  ft. -per- 


(3)  A  point  starts  from  rest  and  moves  with  a  constant  rate  of 
change  of  speed.  Shoio  that  this  rate  is  numerically  equal  to  twice 
the  number  of  units  of  distance  described  in  the  first  second. 


Ans.  We  have  t 

2(8 -Si) 


1  and  ®,  =0;  hence  from  eq.  (5) 


-Si  1 


1  sec 


1  sec. 


,  which  is  numericaUy  equal  to  2(s  —  s,). 


^J^In 


In  an  air-brake  trial,  a  train  running  at  40  miles  an  hour  was 
stopped  in  625.6  ft.  If  the  rate  of  change  of  speed  was  constant 
during  stoppage,  what  was  it  f 

40  y  5*^80 
Ans.  From  eq.  (6),  we  have  for  »  =  0,  s  —  s,  =  625.6,  and  », 


^1"         _ 


(40  X  5280)" 


2  X  625.6  (60  X  60)"  X  2  X625.6 

The  (— )  sign  shows  retardation. 


60  X  60   ' 
=  —  2.75    ft  -per-sec.    per    sec 


U»<1 


A  point  starts  with  a  speed  r,  and  has  a  constant  rate  of 
change  of  speed  —  a.  When  will  it  come  to  rest,  and  what  distance 
does  it  describe  f 


CHAP.  II.]  RATE   OF   CHANGE    OF   SPEED.  31 

Ans.  From  eq.  (3),  when  «  =  0,  we  have  v^  —  at  =  0,    or    t  ~  — .   Prom  eq» 

<"»-'■  =  5 '=-5^- 

\*^  A  point  describes  150  ft.  in  the  first  t/iree  seconds  of  its  motion 
and  50  ft.  in  the  next  two  seconds.  If  the  rate  of  change  of  speed  is 
constant,  when  will  it  come  to  rest  f  When  will  it  have  a  speed  of 
30  ft.  per  sec.  f 

g 
Ans.  From  eq.  (5)  we  have  for  «  —  «j  =  150  and  t  =  S,  150  =  3®,  +  o^  ;    and 

26 
for  5  —  «,  =  300  and  t  =  5,  200  =  5®,  -|-  -^a.    Combine  these  two  equations  and 

we  have  a  =  —  10  ft.-per-sec.  per  sec,  and  v^  =  65  ft.  per  sec.  From  eq.  (3), 
if  B  =  0,  we  have  65  —  10^  =  0,  or  ^  =  6.5  sec.  From  eq.  (3)  we  also  have 
iiv  =  30,  30  =r  65  -  lOt,  oTt  =  3  5  sec. 


v-<?r^ 


A  point  whose  speed  is  initially  30  meters  per  sec.  and  is 
decreasing  at  the  rate  of  40  centimeters-per-sec.  per  sec.,  moves  in  its 
path  until  its  speed  is  240  meters  per  minute.  Find  the  distance 
traversed  and  the  time. 

Ans.  We  have  Vi  =  30  and  ®  =  4  meters  per  sec,  and  a  =  —  0.4  meters- 

per-sec  per  sec.     From  eq.  (6)  s  —  Si  = 7rQ~  —  ^^^^  meters.     From  (3)  we 

—  O.o 
have  4  =  30  -  0.4t,  or  t  —  65  sec. 

LJi&fA  point  has  an  initial  speed  of  Vi  and  a  variable  rate  of 
change  of  speed  given  by  +  kt,  where  k  is  a  cotistant.  What  is  the 
speed  and  distance  described  at  the  end  of  a  time  t  f 

dv  kfi 

Ans.  From  eq.  (9)  we  have  a  =  -—  =  kt,  and  integrating,  v  =  — -  -j-  C.    If, 

(it  i 

kt'* 
when  t  =-0,  we  have  b  =  «, ,  we  obtain  G  =  Vi ,  and  hence  v  =  v^  ~\ — —. 

kPdt  kt^ 

From  eq.  (8)  ds  =  vdt  =  v.dt  H k-.     Integrating,  s  =  ®  /_)_-_  -j-  C.     If, 

a  6 

kt^ 
when  ^  =  0,  we  have  s  =  0,  we  obtain  C  =  0,  and  hence  s  —  v,t-{-  -^-. 

6 

(9)  A  point  has  an  initial  speed  of  60  ft.  per  sec.  and  a  rate  of 
change  of  speed  of  +  40  ft.-per-sec.  per  sec.  Find  the  speed  after  8 
sec.  ;  the  time  required  to  traverse  300  ft.;  the  change  of  speed  in 
traversing  that  distance  ;  the  final  speed. 

Ans.  From  eq.  (3)  we  have  ®  =  60  +  40  X  8  =  380  ft.  per  sec.     From  eq, 

(5)  we  have  300  =  60f  -\-  20<^  or  t  =±y— --  =  +  2.65  sec  or  -  5.65  sec. 

The  first  value  only  applies.  _ 

From  eq.  (3)  we  have  v  -  v,  -  40t  -  20(±  V69  -  3)  =  -f  106  ft.  per  sec.  or 
—  226  ft.  per  sec.     The  first  value  onlv  applies. 

We  have  for  the  final  speed  «  =  +  166  ft.  per  sec.  or  —  166  ft.  per  sec. 
The  first  value  only  applies. 

That  is,  the  point  starts  from  A  with  the  speed  v^  =  60  ft.  per  sec.  and  de- 
scribes the  path  AB  =  300  ft  in  ^  =  2.65  sec,  

the  speed  at  B  being  v  =  166  ft.  per  sec.  ~*" 

In  order  to  interpret  the   negative  values        ^    300/?7 
obtained,  we  observe  that  v  =z  —  166   ft.    per     'A  v^-^  +  eo 
sec.  means  that  the  point  moves  in  the  opposite  t  -  2.65 

direction. 


KINEMATICS— GENERAL   PRINCIPLES.  [CHAP.   II. 


f-5.65 


Let  tlie  point  then  start  from  B  in  the  opposite  direction  with  the  speed  «.  = 
—  166  ft.  per  sec.     Then  from  eq.  (3)  we  have  «  =  —  166  -|-  40^.     We  see  that 

when  t  =  4.15  sec,  v  --  0,  and  thejjoint 
has  passed  to  some  point  P,  where  the 
speed  is  zero.  This  point  is  the  turning- 
point.  For  t  greater  than  4.15  sec.  v  be- 
comes positive;  that  is,  the  point  moves 
back  towards  B,  and  arrives  at  a  point 
A  where  the  speed  is  ®  =  -f-  60  in  the 
time  given  by  60  =  40^,  or  t  —  1.5  sec. 
The  entire  time  from  B  to  P  and  back  to  J.  is  ^  =  5.65  sec.  This  is  the  time 
given  by  the  negative  value  of  t  in  the  example  ;  that  is,  it  is  the  time  before 
the  start,  during  which  the  point  moves  from  B  to  P  and  back  to  A.  The 
change  of  speed  ®  —  Wi  is  -f-  60  -f-  166  =  +  226,  which  is  the  negative  value  in 
the  example. 

For  the  space  BA  described  between  the  initial  and  final  positions,  we  have 

'^^^^t  or  +^Q  ~  ^jg  -^  5  65  ^  _  300  ft.,  the  (-)  sign  showing  that  the  dis- 

tance  is  on  the  other  side  of  the  origin,  from  the  case  of  the  example. 

We  see  then  that  our  equations  are  general  if  we  have  regard  to  the  signs 
of  v,  Vj ,«,«,,  and  a. 

(10)  If  the  motion  in  example  9  is  retarded,  find  (a)  the  distance 
described  from  the  starting  to  the  turning  point ;  (6)  the  distance  de- 
scribed from  the  starting-point  after  10  sec,  the  speed  acquired  and 
the  distance  between  the  final  and  initial  positions  ;  (c)  the  distance 
described  during  the  time  in  which  the  speed  changes  to  —90  ft.  per 
sec,  and  this  time;  (d)  the  time  required  by  the  moving  point  to 
return  to  the  starting-point. 

Ans.   The  initial  speed  is  v^  =  -f-60  ft.  per  sec,  the  rate  of  change  of  speed 
is  a  =  —  40  ft.-per-sec.  per  sec.     Let  the  time  count  from  the  start  at  A,  so 
that  Si  =  0,  when  ^  =  0,  and  let  dis- 
tances and  motion  from  A  towards  P  9  ~^^:^^.^ 
be  positive. 

(a)  We  have  from  eq.  (6)  for  the 
distance  from  A  to  the  turning-point 
P,  where  v  —  0, 

.       -«,»       -3600        ,    ._^ 
-  =  -j-  45  ft. 


p  17  — 0 


2a 


-80 


(b)  From  eq.  (5)  we  have  for  the  distance  between  the  initial  and  final  posi- 
tions after  10  sec.  AB  =  s  =  v^t-{--at^  =  60  X  10  —  20  X  100  =  -  1400  ft. 
The  minus  sign  shows  distance  on  left  of  A.  The  total  distance  described  is 
then  1490  ft.     The  speed  acquired  is  given  by  eq.  (5)  —  1400  =  ^-^ —  X  10 

OT  V  =  —  340  ft.-per-sec.     The  minus  sign  shows  motion  from  A  towards  B. 

(c)  From  eq.  (6)  the  distance  between  the  initial  and  final  positions,  when 

'_  .  _  ^''  -  •"■'  _  8100  -  3600 

2a      ~         -80 

ft.     The  minus  sign  shows  that  C  is  on  left  of  A.    The  total  distance  described 
from  the  start  is  then   56.25  +  90  =  146.25  ft.     The  time,  from  eq.  (5),  is 

_  56.25  =  ~^^  +  ^\  or  t  =  3.75  sec. 

(d)  The  time  to  reach  the  turning-point,  as  we  have  seen,  is  1.5  sec.  The 
time  to  return  is,  from  eq.  (5),  45  =  — ^  =  1.5  sec.  The  entire  time  to  go  and 
return  is  then  8  sec. 


the  speed  is  —  90  ft.  per  sec,  is  ^C=  «  = 


=  -  56.25 


■CHAP.  II.]  RATE   OF   CHANGE   OF   SPEED.  33 

^     (11)  A  railway  train  runs  at  a  speed  of  20  miles  an  hour,  and  its 

speed  is  increasing  uniformly  at  the  rate  of  14  feet-per-min.  per  min. 

Pind  its  speed  after  li  hours,  and  the  distance  traversed  in  that  time. 

Ans.  14  feet-per-min.  per  min.  =;  9.54  miles-per-bour  per  liour.     From  eq. 

(3)  17  —  20  -|-  9. 54  X  1.5  =  34.3  miles  per  liour.     From  eq.  (6)  the  distance  de- 

.,    ^  .    34  3»  -  20«       ,^  „     ., 

scribed  is  —r- t-^-t-  =  40.7  miles. 

1^^^^  2  X  9.54 

(12)  A  railway  train  moving  with  a  speed  of  50  miles  an  hour 
has  the  brakes  put  on,  and  the  speed  diminishes  uniformly  /w  1 
minute,  when  it  is  found  to  he  20  miles  per  hour.  Find  the  rate  of 
change  of  speed,  and  the  distance  traversed.  Also  the  time  in  which 
it  ivill  come  to  rest  and  the  distance  traversed. 

Ans.  From  eq.  (3),  a  = =  — —  —  1800  miles-per-hour  per  hour. 

60 

From  eq.  (6)  the  distance  traversed  is  5^-- —  =  j^  mile.   In  order  to  come 

—  obOO  1^ 

to  rest,  t  = =  — rKP{h  =  oS  liour  =  100  sec.     The  distance  in   coming  to 

(I       —  loOO      00 

-V*       -2500       25     ., 

(13)  Express  a  rate  of  change  of  speed  of  500  centimeters-per- 
^econd  per  second,  in  terms  of  the  kilometer  and  minute. 

Ans.  18  km.-per-min.  per  min. 

C/^  (14)  The  speed  and  rate  of  change  of  speed  of  a  moving  point  at  a 
certain  moment  are  both  measured  by  10,  the  foot  and  second  being 
the  units.  Find  the  number  measuring  them  when  the  yard  ana 
minute  are  the  units. 

Ans.  200  yards  per  min. ;  12000  yards-per-min.  per  min. 

f^  (15)  What  is  meant  when  it  is  said  that  the  rate  of  change  of  speed 
of  a  point  is  +  10,  the  units  being  foot  and  second  f  If  the  point  were 
moxnng  at  any  instant  at  the  rate  of  7i  feet  per  second,  after  what 
time  would  its  speed  he  quadrupled  f  and  what  distance  would  it  de- 
scribe in  that  time  f 

Ans.  2i-sec.;  42.1875  ft. 

(16)  A  body  describes  distances  of  120  yards,  228  yards,  336  yards, 
in  successive  tenths  of  a  second.  Show  that  this  is  consistent  with 
constant  rate  of  change  of  speed,  and  find  the  numerical  value  if  the 
units  are  a  minute  and  a  yard. 

Ans.  38880000  yards-per-min.  per  min. 

^      (17)  Anoint  is  moving  at  the  rate  of  h  feet  per  sec,  a  quarter  of  a 
'     minute  after  at  the  rate  of  5Q  feet  per  sec.,  half  a  minute  after  at  95 

Jeet  per  sec.     Shoio  that  this  is  consistent  with  a  constant  rate  of 

change  of  speed,  and  find  its  value. 
Ans.  3  ft.-per-sec.  per  sec. 


CHAPTEE  III. 

DISPLACEMENT.    RESOLUTION  AND  COMPOSITION  OF 
DISPLACEMENTS. 

Displacement. — The  total  change  of  position,  measured  in  units 
of  length,  of  a  point,  between  its  initial  and  final  positions,  without 
reference  to  the  path  described,  is  the  linear  displacement  of  the 
point. 

Thus  if  a  point  moves  from  the  position  Ai 
to  the  position  Aa,  the  distance  AiAi  is  the  linear 
displacement,  no  matter  what  the  path  may 
have  been  from  Ai  to  ^2. 
The  term  displacement  always  means  linear  displacement  unless 
otherwise  specified. 

Line   Representative  of  Displacement. — The  displacement  of  a 

{)oint  is  therefore  completely  represented  by  a  straight  line.  The 
ength  of  the  line  gives  the  magnitude  of  the  displacement,  and  the 
direction  as  denoted  by  an  arrow  gives  the  direction  of  the  dis- 
placement. 

Thus  the  straight  line  AiAi  represents  by  its 
length  the  magnitude  of  a  displacement,  and  the 
arrow  shows  that  the  displacement  is  in  the  direction 
from  Ai  to  A2. 

A  displacement  then  has  both  magnitude  and 
direction,  and  such  a  quantity  is  called  a  vector  quantity.    All 
vector  quantities  can  Ije  represented  thus  by  a  straight  line.* 

Relative  Displacement.  —  Since  the  displacement  of  a  point  is 
change  of  position,  it  can  only  be  determined  by  reference  to  some 
chosen  point  of  reference. 

Thus  if  Ai  and  Aa  are  the  initial  and  final  positions  of  a  moving 
point  A,  and  Bi  is  some  chosen  point  of  refer- 

Bjf; Ts/  ence,  we  call  AiA,  the  displacement  of  A  with 

\  /^y  reference  to  Bi ,  because  to  an  observer  at  B, 

\      /      \^  the  point  A  would  move  from  ^1  to  A-i  in  the 

\  /  \.        direction  A,Ai. 

B  Aj         But  to  an  observer  on  the  moving  point  A , 

the  point  B,  would  appear  to  move  from  Bi  to 

Bi ,  and  this  is  really  the  displacement  of  B  relative  to  A. 

That  is,  any  change  in  the  relative  position  of  two  points  A  and 

*  As  we  shall  see  liereafter,  linear  velocity  and.  acceleration,  angular  ve- 
locity and  acceleration,  moment  of  linear  velocity  and  acceleration,  moment  of 
angular  velocity  and  acceleration,  are  all  vector  quantities,  and  the  same  prin- 
ciples apply  to  all  of  them  as  to  displacements, 

84 


CHAP.   III.]  DISPLACEMENTS.    ,  35 

B  may  be  regarded  either  as  a  displacement  of  A  relative  to  Boras 
an  equal  and  opposite  displacement  of  B  relative  to  A. 

Relative  Displacement— Two  Points.— If  then  the  straight  line 
AB  represents  the  displacement  of  a  point  A 
with  reference  to  a  point  JS,  the  equal  and  op-     ^ 
posite  line  BA  represents  the  displacement  of  B 
relative  to  A. 

We  shall  always  denote  therefore  relative 
displacement  by  a  line,   the  length  of  which 
gives  the  magnitude  of  the  displacement,  the 
arrow  its  direction,  and  the  letters  at  the  end  the  two  points.    Thus 
in  the  figure  one  line  gives  the  displacement  of  A  relative  to  JB,  the 
other  gives  the  displacement  of  B  relative  to  A. 

Relative   Displacement — Three   Points —Triangle   of  Displace- 
ments.*—Let  a  moving  point  A  have  the  dis- 

^"s;— ^ ^        placement  AB  with  reference  to  a  point  B, 

/       J><''  \  "     ^^^  ^^  *^®  same  time  let  B  have  the  displace- 

/       r""'''^^^'^     ment  BC  with  reference  to  a  point  C, ,  and  let 

qZ. >5.i ~>c  B^ ,  Bi  be  the  initial  and  final  positions  of  the 

\     ^'^'        point  B. 
v^  Then  it  is  evident  that  since  the  point  has 

*  the  displacement  AB  with  reference  to  B,  and 

at  the  same  time  B  moves  from  B^  to  Bi ,  the  point  moves  from  A  to 
C,  and  AC  is  the  displacement  of  A  with  reference  to  C. 

Conversely,  from  the  preceding  article,  CA  is  the  displacement 
of  C  relatively  to  A. 

Hence,  if  two  sides  of  a  triangle  ABC  taken  the  same  way  round 
represent  the  displacements  of  A  relative  to  B  and  B  relative  to  C 
respectively,  the  third  side  taken  the  opposite  ivay  round  will  repre- 
sent the  displacement  of  A  relative  to  C,  and  taken  the  same  y:>aUr      ,  J 
round,  the  displacement  of  C  relative  to  A.  Ar^^  ^s/r^  i)  r 

This  is  called  the  principle  of  the  triangle  of  displacement^,    xfc--'    ^  ^ 
evidently  makes    no   difference  whether   the  displacements  are 
simultaneous  or  successive.      The  same  principle  holds  true  in. 
both  cases. 

Polygon  of  Displacements. — Let  AB, 
BC  and  CD  be  the  given  displacement  of 

A  relative  to  B,  B  relative  to  C,  and  C    ^3 

relative  to  D.  Then  if  we  lay  off  succes- 
siyely  AB  and  BC  in  given  direction  and 
magnitude,  the  line  AC  gives  the  displace- 
ment of  A  relative  to  C.  If  we  lay  off 
CD,  the  line  AD  gives  the  displacement  of 
A  relative  to  D.  - 

Hence,  if  any  number  of  displacements 
in  the  same  plane  are  represented  by  the 
sides  of  a  plane  polygon,  ABCD,  etc.,  taken  d 

the  same  way  round,  the  line  AD  which  closes  the  polygon  taken  the 
opposite  way  round  will  give  the  magnitude  and  direction  of  the  re- 
sultant displacement  of  the  first  point  relative  to  the  last,  and  taken 
the  same  way  round,  of  the  last  point  relative  to  the  first. 

This  principle  is  called  the  polygon  of  displacements,  and  it  evi- 

*  We  shall  see  hereafter  that  all  the  principles  which  follow  in  this  chap- 
ter relating  to  displacements  hold  equally  good  for  linear  and  angular  veloci- 
ties and  accelerations,  for  the  moments  of  linear  and  angular  velocities  and  ac- 
celerations, and  for  forces  and  moments  of  forces. 


B  C 

\/ 


36 


KINEMATICS — GENERAL    PRINCIPLES.  [CHAP.    III. 


dently  holds  good  whether  the  displacements  are  successive  or 
simultaneous. 

Resolution  of  Displacements. — By  the  application  of  these  prin- 
ciples a  given  displacement  or  any  number  of  given  displacements 
may  be  resolved  into  two  components  in  any  two  given  directions. 
Thus  suppose  the  displacement  of  A  relative 
to  -B  to  be  given  by  the  line  AB.  We  can  re- 
solve this  displacement  into  components  in  any 
two  directions  given  by  the  arrows  a  and  6,  by 
drawing  lines  from  A  and  B  parallel  to  these 
directions  till  they  intersect  at  some  point  C. 
Then  the  displacements  AC  and  CB  taken  the 
other  way  round  are  the  component  displacements 
of  AB  in  the  required  directions. 
If  any  number  of  displacements  AB,  BC,  CD,  etc.,  ^ 
are  given,  we  have  the  resultant  displacement  AD, 
and  this  displacement  can  be  resolved  as  before  into 
any  two  directions  required. 

Rectangular  Components.— When  a  displacement 
is  thus  resolved  into  two  directions  at  right  angles, 
the  components  are  called  rectangular  components. 
Unless  otherwise  specified,  when  we  speak  of  the  components  of 
any  displacement,  the  rectangular  components  are  always  under- 
stood. 

The  Component  of  the  Resultant  is  equal  to  the  Algebraic  Sum 

of  the  Components  of  the  Displacements.  — It  is  evident  that  the 

resultant  of  any  two  given  displacements  is  equal 

_^B  to  the  algebraic  sum  of  their  components  in  the 

direction  of  the  resultant. 

For  if  AC  and  CBaxe  the  given  displacements, 
the  resultant  AB  is  equal  to  the  sum  of  the  com- 
ponents AD  and  DB. 

So  also  for  any  number  of  displacements,  the 

resultant  AE  is  equal  to  the  algebraic  sum  of  the 

components  of  the  displacements  in  the  direction 

otAE. 

The  component  in  any  given  direction,  of  this 
resultant  itself,  is  then  equal  to  the  algebraic  sum 
of  the  components  of  the  displacements  in  the 
same  direction. 

Thus  the  projection  of  the  resultant  AE  upon 
the  line  OP,  or  the  component  of  AE  in  the  direc- 
tion OP,  is  the  algebraic  sum  of  the  components 
Y  of  AB,  BC,  CD  and  BE  in 

the  direction  OP. 

Components    not    in    the 
Same  Plane. — The  same  principles  apply  for 
components  not  in  the  same  plane. 

Thus  if  OA  is  a  given  displacement,  and  we 
draw  AB  perpendicular  to  the  plane  of  XZ 
meeting  it  at  B,  we  have  the  components  OB 
and  BA. 

Again,  we  can  resolve  OB  into  the  compo- 
nents OC  and  CB.     The  components  then  are  OC,  CB  and  BA. 

Sign  of  Components  of  Displacement. — A  sign  of  (4-)  or  (— )  pre- 
fixed to  the  magnitude  of  a  component  displacement  indicates 
direction.    Thus  for  three  rectangular  axes  OX,  OY,  OZ,  a  com- 


CHAP.   III.] 


DISPLACEMENTS. 


37 


ponent  displacement  in  the  directions  OX,  OY,  OZ  is  positive,  and 
in  the  opposite  directions  negative. 

If  polar  co-ordinates  are  used,  the  com- 
ponent displacement  along  the  radius  vector 
is  positive  when  away  from  the  pole,  nega- 
tive when  towards  the  pole. 

Evidently,  then,  we  measure  angles  in 
the  plane  XY  from  OX  around  towards 
0Y\  in  the  plane  YZ  from  OY  around 
towards  OZ;  in  the  plane  ZX  from  OZ 
around  towards  OX,  as  shown  by  the 
arrows  in  the  figure. 

Analytical  Determination  of  the  Resultant  for  Concurring  Dis- 
placements,— When  the  line  representatives  meet  in  a  point  they 
are  called  concurring.  All  displacements  of  a  single  point  must  be 
concurring.  The  magnitude  and  direction  of  any  number  of  such 
concurring  component  displacements  being  given,  to  find  expres- 
sions for  the  magnitude  and  direction  of  the  resultant. 

(a)  When  there  are  Two  Given  Components.  — Let  OB  —  di  and 

BC  =  di  be  two  component  displacements 

making  the  angle  a  and  in  the  indicated 

directions. 

Then  the   resultant  is  OC  =  dr  and  is 

given  at  once  in  magnitude  from  the  tri- 
angle OBC, 

dr'  =  {di  +  di  cos  ay  +  (di  sin  ay  —  di"  +  2d,di  cos  a  +  d-y.    .     (1) 

The  resultant  dr  makes  with  d,  an  angle  a,  given  by 


cos  a  = 


di  +  di  cos  aa 

dr 


(2) 


(6)  When  there  are  Any  Number  of 
Components  in  any  Given  Direction. — 
Take  three  rectangular  axes,  OX,  OY, 
OZ,  through  the  point  O,  and  let  the 
component  displacements  di,  di,  da, 
etc.,  make  the  angles  «r,,  /S,,  y  -^  a^,  /S^, 
r-!,  etc.,  with  the  axes  of  X,  Y,  Z respect- 
ively. 

Then  we  have  for  the  sum  of  the 
components  in  the  direction  of  X,  Y, 
and  Z, 

dx  =  di  cos  a-i  +  di  cos  a,  +  . . .  =  .2d cos  a; 
dy  =  di  cos  A  +  di  cos  A  -f-  .  . .  =  2dcos/3; 
dz  =  di  cos^i  +  di  cos  Vi  +  . .  .=  ^dcosy.  ) 

In  these  summations  we  must  take  components  with  their  proper 
signs  as  directed  in  the  preceding  Article. 

We  have  then  for  the  magnitude  of  the  resultant  dr,        naA-^ 

dr  =  Vdx'  +  dy'  +  dzK    .     .    .  J^-P-r.    .    (4) 

This  resultant  makes  with  the  axes  angles  a,  6,  c,  respectively, 
given  by 

"'(5) 


d^ 


r       dii  dz 

cos  a  =  3- ,    cos  6  =  -="-,    cos  c  =  --. 

dr  Or  <*r 


^i 


"38  KINEMATICS — GENERAL   PRINCIPLES.  [CHAP.    III. 

The  projections  on  the  planes  XY,  YZ,  ZX  make  angles  with  X, 
Y,  Z,  respectively,  whose  tangents  are 

Uy  dz^  dx  ,„, 

dx  CLy  CLz 

Cor.  1.  If  all  the  displacements  are  in  one  plane,  make  dz  =  0. 
Then  ^_^_ 

dr  =  \/dx'  +  dy' (7) 

cos  a  =  ~,    cos  b  =  ~\ (8) 

dr  dr 

and,  since  cos  b  =  sin  a, 

tana  =  ^ (9) 

dx 

Cor.  2.  When  there  are  hut  two  displacements,  di  and  d,  take 
OX  corresponding  with  d, .   Then  dx  =  di  +  d  cos  a,dy  —  d  sin  or,  and 
we  have  at  once  equations  (1)  and  (2). 
^^       Cor.  3.  If  the  two  displacements  are  in  the  same  direction  or  in 
j/V  "^     opposite  directions,  a  =  0  or  180°,  and 


dr  ^=di  ±  d. 

That  is,  the  resultant  is  the  algebraic  stun  of  the  displacements. 
I  ,^/-^CoR.  4.  If  the  two  displacements  are  at  right  angles,  «  =  90°  and 

ji/sA^"'  d  d  d 

dr  =  Vdi'  +  d',     sin  a  =  ^— ,     sin  &=-_—=  cos  a,     tan  a  =  -r- 

dr  dr  di 


Cor.  5.  If  the  two  displacements  are  equal,  di  =  d  and 
dr'  =  2d*(l  +  cos  a)  =  4d'  cos'  ^, 


./: 


hence  dr  =  2d  cos  ^. 

. ,              .            d  sin  a        .    a     .  a 

I       Also,        sin  a  = =  sm  — ,    hence    a  =  — 

ri^  /-  «  ,  «  2  2 

j^"  (  2dcos- 

CoR.  6.   If  the  two  displacements  are  equal  and  a  =  120",  then 
dr  =  d,  and  a  =  60°. 


EXAMPLES. 

(1)  A  wheel  whose  radius  is  r  rolls  on  a  horizontal  plane  until  it 
turns  through  a  quarter  revolution.  Find  the  displacement  of  the 
point  of  the  wheel  initially  in  contact  with 
the  plane  relative  to  the  point  diametric- 
ally opposite. 

Alls.  Let  C  represent  the  initial  point  of  con- 
tact of  tlie  plane,  A  the  point  of  tlie  wheel  in  con- 
tact  with  C  in  the  plane,  B  the  point  of  the  wheel 
diametrically  opposite  A. 

The  displacement  of  B  with  reference  to  C  is 


CHAP.    III.]  DISPLACEMENTS — EXAMPLES.  39 

given  by  BiBi. 


y^'A 


The  displacement  of  A  with  reference  to  Cis  given  by  AxA^, 
and  therefore  the  displacement  of  C  with  reference  to 
A  is  given  by  A^Ai.  We  have  then,  by  laying  oS 
these  displacements,  the  displacement  ot  A  relative 
to  B  equal  t(?  2£j/^ ,  making  an  angle  of  45°  with 
the  verticUh;^!^.' 

*tf|L-rfncZ  the  displacement  of  the  end  of 
the  frninute-hand  with  reference  to  the  end  of 
the  hour-hand  of  a  clock,  between  3  and  3.30 
o'clock,  the  length  of  each  hand  being  r. 

Ans.    The    displacement   of   the   minute-hand   with   reference  to  XII  is 
MiMi ,  and  of  the  hour-hand  with  reference  to  XII,  HiH, . 
Therefore  the  displacement  of  XII  with  refer- 
ence to  ZT  is  HiHi .     Laying  off  these  displace- 
ments, we  have 


XliSi 


MR=  r  -/e  -  3 cos  15°  —  4  sin  15°, 

and  for  the  angle  of  MH  with  the  vertical 
2  sin2  7^° 


sm^JfXII 


J 


V'6  —  2  cos  15°  —  4  sin  15° 


(^ 


(3)  A  river  flows  in  a  direction  N.  33°  E.,  and  a  boat  headed  directly 
axiross  at  right  angles  to  the  current  reaches  a  point  on  the  other 
shore  from  which  the  starting-point  is  found  to  bear  S.  3°  W.  If  the 
distance  from  the  starting-point  is  500  ft.,  how  far  has  the  boat  been 
carried  by  the  current,  andivhat  is  the  distance  across  the  river  if  the 
banks  are  parallel  f 

Ans.  The  course  of  the  boat  makes  an  angle  of  30°  with  the  bank.  The 
distance  across  is  250  ft. ;  the  distance  parallel  to  the  bank  433  ft. 

I  ^^tA^  ■   (4)  A  point  A  moves  30  ft.  in  a  given  direction  relative  to  a  fixed 
Jj       point  O.     Another  point  B  moves  relative  to  O  40  ft.  in  a  direction 
Jr  at  right  angles  to  A's  motion.    Find  the  displacement  of  A  relative 

to  B. 

Ans.  50  ft.  in  a  direction  inclined  to  A'&  motion  by  an  angle  whose  sin 


JWhTwo  points  move  in  the  circumference  of  two  circles  of  radius 
r  arid  2r  respectively.  Both  start  from  the  point  of  contact.  One 
moves  through  an  angle  of  n  radiants,  the  other  through  an  angle  of 

Tt  * 

—  radians.    Find  the  displacement  of  either  relative  to  the  other. 

Ans.  If  one  point  moves  through  the  angle  Tt  in  the  small  circle  and  the 
other  through  —  in  the  large,  the  displacement  is  2r  4/5 ,  making  an  angle 

with  the  vertical  whose  tangent  is  2. 

If  one  point  moves  through  the  angle  it  in  the  large  circle  and  the  other 

through  -^  in  the  small,  the  displacement  is  rv'26  and  it  makes  an  angle  with 

the  vertical  whose  tangent  is  5. 

IJ^-Jn  the  preceding  example  let  the  radius  of  the  small  circle  be 
Ti  ahd  of  the  large  circle  r,.    Find  the  displacement  as  before. 


40 


KINEMATICS — GENERAL  PRINCIPLES.  [CHAP.    IIL. 


^ 


if 


Ans. 


'l/4ri''  +  4r2n  +  3ri»  and  tangent 


2u  +  r. 


Ti 


V4ra'  +  4»"a»*i  4-  Sti"  and  tangent 


2ra  +  ri 


ri 


(7)  TTie  displacement  of  A  relative  to  B  is  a  distance  a  towards 
the  south,  and  relative  to  C,  c  towards  the  west.  If  C  is  initially  a 
distance  b  south  of  B,  find  the  final  position  of  C  relative  to  B. 

Ans.  We  have  CB  for  the  displacement  of 
C  relative  to  B.  Therefore  if  Ci  is  the  initial  po- 
sition, C-,  will  be  the  final  position,  CjCo  being 
equal  to  CB.  Hence  the  distance  of  C's  final 
position  from  B  is  Bd  =  |/(6 -f  ay^^\  and 
the  direction  from  B  to  Cj  is  east  of  south  an 


angle  whose  tangent  is 


6  + a" 


(8)  A's  displacement  relative  to  B  is  a  to  the  west.  C^s  displace- 
ment relative  to  B  is  c  in  a  direction  30°  west  of  south.  Find  the 
displacement  of  A  relative  to  C. 

Ans.  Displacement  is  |/a'  —  ac-{-  c" 
.     2a  —  c 


Tangent  of  the  angle  with  the  merid- 
If  this  is  positive,  the  angle  is  west  of  north;  if  negative,  east 


ci/3 
of  north. 

»''  (9)  Two  trmms  A  and  B  start  from  the  same  point  and  A  runs 
60  miles  northand  50  miles  northeast.  Find  the  displacement  of  B 
relative  to  A. 

Ans.  43.104  miles  in  a  direction  34°  52'  south  of  east. 

'   (11)  A  point  undergoes  two  component  displacements,  100  feet  W. 
30°  S.  and  30  ft.  N.     What  is  the  resultant  f 
/    Ans.  88.88  ft.  13°  south  of  west.  -> 

(12)  Three  component  displacements  have  magnitiides  represented 
by  1,2  and  3  and  directions  given  by  the  sides  of  an  equilateral  tri- 
angle, taken  the  same  way  around.    Find  the  magnitude  of  the  re- 
sultant. 
,  Ans.     4/3. 

"p^  Two  railway  trains  run  on  parallel  roads,  one  5  miles  north, 
thS  other  Q  miles  south.  Find  the  displacement  of  the  last  relative 
to  the  first. 

Ans.  11  miles  south. 


(14)  Two  railway  trains  run,  the  one  northeast  a  distance  d,  the 


other  southeast  the  same  distance.     Find  the  displacement  of  the 
first  relative  to  the  last. 

Ans.  d  4/2]  direction  north. 

)/  (15)  A  point  undergoes  three  component  displacements,  N.  60°  E. 
40  ft.,  S.  50  ft ,  W.  30°  N.  GO  ft.    Find  the  resultant. 

wAns.  10  i''3  ft.  W. 

"Sl^  "^  ^^^  ^^^9(^^^^  -^C  of  a  quadrilateral  ABCD  is  produced  to 
E,  J^^Jiat  CE  is  equal  to  AC,  show  that  AE  is  the  resultant  in  the 
difkctior  AC  of  the  displacements  AD,  DB,  BC  and  AC. 

\  V  wl7)  ABCD  is  parallelogram.    E  is  the  middle  point  of  AB.  FinxL 


y. 


CHAP.   III.]  DISPLACEMENTS — EXAMPLES.  41 

the  components^  in  the  directions  of  AB  and  AD,  of  a  displacement 
which  has  the  direction  and  half  the  magnitude  of  the  resultant  of 
component  displacements  represented  by  AC  and  AD. 
Ans.  AE  and  AB. 

(18)  Prove  that  the  resultant  of  two  equal  displacements  of  mag- 
nitude a,  inclined  60%  is  equal  to  the  resultant  of  a  and  2a  inclined 
120°. 


/'' 


:  .   ,-  (19)  Prove  that  if  two  component  displacements  are  represented 

f'      by'iwo  chords  of  a  circle  drawn  from  a  point  P  in  the  circumference 

perpendicular  to  each  other,  the  resultant  is  represented  by  the  diame- 

',  ter  through  the  point. 

^j       (20)  To  an  observer  in  a  balloon  the  starting -ppint  bears N.  20°  E.,wjt 
JAand  is  depressed  30  below  the  horizontal.    A  point  at  the  same  level 

as  the  starting-point  and  10  miles  from  it  is  vertically  below  him.    ^j. 

What  are  the  component  displacements  of  the  balloon  with  reference 

to  the  starting-point  f 

Ans.  9.39  miles  south,  3.43  miles  west,  and  5.77  miles  up.  v-J 

rA^^\jJ  (21  \  A  point  has  the  component  displacements  in  the  same  plane, 
'  q~=  'iOft.,  d-i  =  50  ft.,  da  =  60  ft.,  making  the  angles  with  X  and  Y, 

a,  =  60°,  /?  =  150° ;  a^  =  120°,  /^^  =  30° ;  a,  =  120°,  ^a  =  150°.  Find  the 

resultant  displacement. 

Ans.  (?x  =  4-  20  -  25  -  30  =  -  35  ft.; 

dy^-  34.64  +  43.3  -  51.96  =  -  43.3  ft.; 


dr  =  V^x'  +  dy^  =  55.67  ft. 

cos  a  =  ^  =  g^,  or  a  =  128°  57'  17" ;    cos  6  =  ^  =^^.  OTb  =  141°  3'  43". 
dr      55.67  dr      55.67 

(22)  A  point  has  the  component  displacements  d,  =  40  ft.,  d,  =  50 
ft.,  da  =  QOft.,  making  with  X,  Y,  Z,  the  angles  a,  =  60°,  ft,  =  100°, 
r,  obtuse ;  a^  =  100°,  fh  —  60°,  y^  acute  ;  a^  =  120°,  A  =  100°,  y^ 
acute.    Find  the  resultant  displacement. 

Ans.  We  find  the  angles  y,  (page  12)  by  the  equation 
cos'  y  =  —  cos  (a  4-  0)  cos  (a  —  ft). 
Hence  yi  =  148°  2'  31".7,      y-,  =  31°  57'  28".3,      y^  ^  31°  57'  38".3. 

(Z^  =  -f  30  -  8.6824  -  30  =  -  18.6824  ft. 

dy  =-  6.946  +  25  -  10.419  =  +  7.635  ft. 

dz  =-  33.937  +  42.431  +  50.907  =  +  59.391  ft. 

-1  Q  AQ04, 

dr  =  VdJ+'dy^  +  d,»  =  63.73  ft.  cos  a  =  — gg^^g— .  or  a  =  107°  19*  36"; 
c,s6  =  i|:|-,  or  6  =  83°  0'  33";  cos  c  =  "tg|^,  or  c  =  18°  46'  43'. 


^ 


fy)         ^loy^ 


:v 


A 


CHAPTEE  IV. 


VELOCITY.     RESOLUTION  AND  COMPOSITION  OF 
VELOCITIES. 


BKCTANQULAK  COMPOKENTS  OP  VELOCITY.     SIGN  OP  COMPONENTS  OP  VELOCITT. 
ANAIiTTIC   DETERMINATION  OP  RESULTANT  VELOCITY. 

Mean  Velocity. — The  distance  described  by  a  moving  point  per 
unit  of  time  we  have  called  the  mean  speed  of  the  point  (page  15). 

The  displacement  (page  34)  of  a  point  per  unit  of  time  we  call 
the  mean  linear  velocity.  Therefore  the  number  of  units  in  the  dis- 
placement of  a  point  in  any  given  time,  divided  by  the  number 
of  units  in  that  time,  gives  the  number  of  units  of  mean  linear 
velocity — or  the  magnitude  of  the  mean  linear  velocity — while  the 
direction  of  that  velocity  is  the  same  as  that  of  the  displacement. 

The  term  velocity  always  signifies  linear  velocity  imless  other- 
wise specified. 

Mean  speed,  then,  is  mean  time-rate  of  distance  described  (page 
15).    Mean  velocity  is  mean  time-rate  of  displacement. 

Thus  if  a  point  moves  in  any  path  P^AP^  from 

the  initial  position  Pi  to  the  position  P2  in  the  time  t 

Py^ — -p>P2     units,  the  magnitude  of  the  mean  speed  is  given  by 

f^^  £■ —^ ?,  while  the  magnitude  of  the  mean  ve- 

''i  1     -A     •    ^-         u    displacement  PjP^        ■,  ..      ,. 

locity  IS  given  by  - — ^ — -,  and  its  direc- 
tion by  the  direction  of  P^P-t. 

If  a  point  moves  with  uniform  speed  in  a  circle  of  radius  r  units  and  the 

27tr 
time  of  revolution  is  t  units,  the  mean  speed  is  —7-  units  of  speed.     But  the 

mean  velocity  in  the  time  of  one  revolution  is  zero,  because  the  disi)lacement 

TtT 

in  that  time  is  zero.     Again,  the  mean  speed  in  one  half  a  revolution  is  —    or 

2' 

units  of  speed,  as  before,  but  the  mean  velocity  for  the  same  time  has  for 

its  magnitude  77  or  —  units  of  velocity,  and  the  direction  is  that  of  a  ^iameter 

■5'  * 

from  the  initial  to  the  final  position  of  the  point. 

Instantaneous  Velocity. — The  limiting  magnitude  of  the  mean 
speed  when  the  interval  of  time  is  indefiiiitely  small  we  have  called 

42 


CHAP.    IV.]  VELOCITT.  43 

(page  15)  the  instantaneous  speed.  In  the  same  "way,  the  limiting 
magnitude  and  direction  of  the  mean  velocity  when  the  interval  of 
time  is  indefinitely  small  is  called  the  instantaneous  velocity. 

The  terms  speed  and  velocity  should  always  be  understood  to 
mean  instantaneous  speed  and  instantaneous  velocity  unless  other- 
wise specified. 

Using-  the  terms  thus,  we  see  that  when  t  is  in- 
definitely small,  distance  described  and  displacement 
coincide  and  hence  the  speed  at  Pi  is  the  magni- 
tude of  the  velocity  at  P, ,  while  the  direction  of  this 
velocity  at  any  instant  is  that  of  the  tangent  to  the  Pi 
path  at  that  instant.  Velocity  is  directed  speed.  Speed  is  magni- 
tude of  velocity. 

Unit  of  Velocity. — Since,  then,  the  magnitude  of  the  velocity  at 
any  instant  is  the  speed  in  the  direction  of  the  velocity  at  that  in- 
stant, it  follows  that  the  unit  of  velocity  is  the  same  as  the  unit  of 
speed  (page  15j,  or  one  unit  of  length  per  vmit  of  time,  as,  for 
instance,  one  foot  per  second.  For  this  reason  we  have  used  (page 
25)  the  letter  v  for  the  niimeric  for  speed. 

Uniform  Velocity. — When  the  velocity  has  the  same  magnitude 
and  direction  whatever  the  interval  of  time,  it  is  uniform. 

Uniform  velocity,  then,  is  necessarily  uniform  speed  in  a  straight 
line  in  a  given  direction.  The  velocity  in  such  case  is  the  same  as 
the  mean  velocity  for  any  interval  of  time. 

Variable  Velocity. — When  either  the  magnitude  or  direction  of 
the  velocity  changes  it  is  variable. 

When  the  magnitude  alone  changes,  we  have  variable  speed  in 
a  straight  line.  When  the  direction  only  changes,  we  have  uniform 
speed  in  a  curved  line.  When  both  change,  we  have  variable  speed 
in  a  curved  Une.    In  all  these  cases  the  velocity  is  variable. 

Thus  we  can  speak  of  a  point  moving  in  a  circle  with  uniform  speed,  but 
we  cannot  speak  of  a  point  moving  in  any  curve  at  all  with  uniform  velocity. 
If  the  velocity  is  uniform,  the  path  must  be  straight,  as  well  as  speed  constant. 

A  point  can  be  projected  with  the  same  speed  in  many  different  directions, 
but  we  cannot  speak  of  the  same  velocity  in  different  directions.  A  change  of 
direction  is  a  change  of  velocity  whether  the  speed  changes  or  not. 

Velocity  a  Vector  ftuantity.— Since  velocity  is  speed  directed,  or 
time-rate  of  displacement,  it  has  not  only  sign  and  magnitude  like 
speed  (page  15),  but  also  direction  like  displacement,  and  is  there- 
fore a  vector  quantity. 

Line  Representative  of  Velocity.— Velocity,  then,  can  be  repre- 
sented like  displacement  (page  34)  by  a  straight  line.  The  length 
of  this  line  represents  the  magnitude  of  the  velocity,  and  the  direc- 
tion as  denoted  by  an  arrow  gives  the  direction. 

Thus  the  straight  line  A.Aj  represents  by  its 
length  the  magnitude  of  a  velocity,  and  the  arrow 
shows  its  direction. 

Triangle  and  Polygon  of  Velocities. — The  prin- 
ciples therefore  of  page  35  hold  good  for  velocities 
as  well  as  displacements.     We   have  then  the 
triangle  and  polygon  of  velocities. 

Resolution  and  Composition  of  Velocities.— We  can  also  combine 
and  resolve  velocities  in  the  same  way  precisely  as  displacements, 
and  the  principles  of  page  36  apply  here  also. 


44 


KINEMATICS — GENERAL   PRINCIPLES.  [CHAP.    IV. 


^ 


Eectangular  Components  of  Velocity.— If  a  point  moves  in  any 
path  from  P,  to  P  in  the  time  t,  the  dis- 
placement is  the  chord  PiP,  and  the  mean 

'"                                 7     V    •            chord  P,P 
velocity  18  v^ — t — . 

->vx  If  X,  and  Pi  are  the  co-ordinates  of  P,,  ^ 

and  X  and  y  of  P,  then  the  horizontal  and 
vertical  components  of  the  mean  velocity 


4- 


Wi 


X    V  are 


c 


VI 


J)t 


t'% 


If  the  time  is  indefinitely  short,  we  have  in  the  notation  of  the 
Calculus,  for  the  instantaneous  velocities, 

dx  dy 


nJL,  I'P,  ^Jio. 


rx  = 


dt 


Vy  = 


dt' 


Sign  of  Components  of  Velocity. — We  see  that  if  Vx  is  directed 
towards  the  right,  we  have  in  the  1st  and  4th  quadrants  x  numeric- 
ally greater  than  x,  and  both  are  positive.  In  the  2d  and  3d  quad- 
rants, if  Vx  is  directed  towards  the  right,  x,  is  numerically  greater 
than  X  and  both  are  negative.  In  all  quadrants,  then,  Vx  will  be 
positive  when  directed  towards  the  right.  In  the  1st  and  2d  quad- 
rants, if  Vy  is  directed  upwards,  y  is  numerically  greater  than  yi 
and  both  are  positive.  In  the  3d  and  4th  quadrants,  if  Vy  is  directed 
upwards,  yt  is  numerically  greater  than  y  and  both  are  negative. 
In  all  quadrants,  then,  v,/  will  be  positive  when  directed  upwards. 

We  have  then  the  following  general  rule  for  the  signs  of  the 
components  of  the  velocity  in  any  quadrant : 

If  the  direction  of  the  line  representative  is  toivards  the  right, 
Vx  /  *'  positive  ;  if  towards  the  left,  Vx  is  negative.  If  upwards,  Vy  is 
positive;  if  downwards,  negative. 

The  sign  then  as  applied  to  component  velocities  indicates  direc- 
tion of  motion.  For  rectangular  co-ordinates  (+)  signifies  towards 
the  right  or  upward,  (— )  towards  the  left  or  downward. 

For  three  rectangular  axes  OX,  OY,  OZ,  let  a  point  Pbe  given 
by  the  space  co-ordinates  x,  y,  z.     Let  the 
velocity  v  of  the  point  make  the  angles  «, 
fi  and  r  with  the  axes  of  X,  Y  and  Z. 
Then  we  have 


/ 


X 


Vx  —  V  cos  a,      Vy  =V  cos  /J,      Vz=V  COS  y 


y\ 


/z 


.z 


for  the  components  in  the  direction  of  the 

X     axes.     When  the  angles  a^  (i^  y  are  acute 
or  less  than  90°,  these  components  are  posi- 
tive ;  when  the  angles  are  obtuse  or  more 
than  90%  these  components  are  negative. 
Therefore,  as  before,  Vx  towards  the  right  is  positive,  towards, 
the  left  negative,  Vy  upwards  is  positive  and  downwards  negative, 
and  Vz  in  the  direction  OZ  is  positive,  in  the  opposite  direction, 
negative. 

If  polar  co-ordinates  are  used,  the  component  velocity  along  the 
radius  vector  must  he  taken  as  positive  when  it  acts  away  from  the 
pole,  and  negative  when  it  acts  toioards  the  pole.* 

*  Evidently,  then,  we  measure  angles  in  the  plane  XYfrom  OX  around  to- 
wards 07";  in  the  plane  YZ  from  OF  around  towards  OZ;  in  the  plane  ZX 
from  OZ  around  towards  OX,  as  shown  by  the  arrows  in  the  figure. 


•CHAP.    IV.]  VELOCITY — EXAMPLES.  45 

Analytical  Determination  of  the  Resultant  for  Concurring  Veloci- 
ties.—When  the  line  representatives  meet  in  a  point  they  are 
dialled  concurring.  We  have  then  the  same  expressions  for  the 
magnitude  and  direction  of  the  resultant  of  any  number  of  concur- 
ring component  velocities  as  for  displacements  (pages  36  and  37). 
We  have  only  to  substitute  v  in  place  of  d. 

EXAMPLES. 

(1)  To  a  man  driving  easttvard  with  a  speed  of  4  miles  an  hour, 
th3  ivind  blows  apparently  from,  the  north,  hut  ivhen  he  doubles  his 
speed  the  wind  appears  to  blow  from  the  northeast.  Find  the  real 
direction  and  velocity  of  the  ivind. 

Ans.  The  wind  blows  from  the  northwest  with  a  velocity  of  4  1^2  miles  an 
hour. 

*'  (2)  A  point  moves  in  a  straight  line  from  A  to  B,  QOft.  W.  30°  S., 
in  10  sec,  and  thence  in  a  straight  line  to  C,  30  ft.  N.,  in  20  sec.  Find 
the  mean  speed  and  the  mean  velocity. 

Ans.  The  length  of  path  is  90  ft.  traversed  in  30  sec,  or  mean  speed  is  3  ft. 
per  sec.  The  displacement  is  30  |/3  ft.  W.,  or  the  mean  velocity  is  V3  ft.  per 
sec.  W.  "" 

^  (3)  A  ship  sails  N.  30°  E.  at  10  miles  an  hour.  Find  its  easterly 
velocity  and  its  northerly  velocity. 

Ans.  5  miles  an  hour;  54^3  miles  an  hour. 

(4)  A  river-ir^iii&ixxBXd  has  a  current  of  5  miles  per  hour,  and 
<x  boat  capable  of  making  10  miles  an  hour  in  Stillwater  is  to  go 
■Straight  across.    In  what  direction  m,ust  the  boat  be  steered  f  -        >■< 

Ans.  Up  stream  in  a  diiection  inclined  60°  with  the  bank.  _  , 

(5)  Find  the  vertical  velocity  of  a  train  moving  up  a  1-per-cent        1^^ 
gradient  at  a  speed  of  30  miles  per  hour.  /!?«£•♦ 

Ans.  0.3  mile  per  hour.  ff        /j 

(6)  A  man  travelling  4  miles  per  hour  east  finds  the  wind  to  come 
from  the  southeast.  When  he  stands  still  it  shifts  5°  to  the  south. 
Find  its  velocity.  /o° 

Ans.  32.52  miles  per  hour  N.  40°  W. 

l^-'tf)  A.  point  moving  with  uniform  speed  in  a  circle  of  radius  30 
ft.  describes  a  quadrant  in  10  sec.  Find  the  mean  speed,  the  mean 
velocity,  the  instantaneous  speed  and  the  instantaneous  velocity. 

TtT 

Ans.  The  length  of  path  described  in  10  sec.  is  -5-  =  47.12  ft.  The  mean 
speed  is  then' 4. 712  ft.  per  sec,  and  since_this  is  uniform  it  is  also  the  instan- 
taneous speed.  The  displacement  is  r  |/2  =  42.42  ft.  at  an  angle  of  45°  with 
the  diameter  through  the  starting-point.  The  mean  velocity  is  then  4.242  ft. 
per  sec.  in  the  same  direction.  The  instantaneous  velocity  is  at  any  instant 
tangent  to  the  circle  at  the  point  at  that  instant,  and  equal  in  magnitude  to  the 
instantaneous  speed,  or  4.712  ft.  per  sec. 

V  \\aj«)  J.  man  loalks  at  the  rate  of  4  miles  per  hour  in  a  rain-stoi-m, 
ys<md  the  drops  fall  vertically  tvith  a  ^peed  of  200  ft.  per  sec.    In  what 
'     d/irection  will  they  seem  to  him,  to  fall  f 
Ans.  Inclined  1°  40. 8  to  the  vertical. 


v/  46  ^  KINEMATICS — GENERAL   PRINCIPLES.  [CHAP.   IV. 

\.--^  A  ship  sails  east  with  a  speed  of  12  miles  an  hour,  and  a  shot 
is  fired  so  as  to  strike  an  object  which  bears  N.E.     If  the  gun  gives 
the  shot  a  mean  horizontal  velocity  of  90  ft.  per  sec,  towards  what 
point  of  the  compass  must  it  point  f 
Ans.  N.  37°  3'  E. 

~~  (10)  A  man  6  ft.  tall  ivalks  at  the  rate  of  4  miles  per  hour  directly 

away  from  a  lamp-post  10  ft.  high.    Find  the  velocity  of  the  ex- 
tremity of  his  shadow. 

Ans.  10  miles  per  hour  in  the  direction  he  is  walking  (see  Ex.  14,  page  19). 

(11)  Two  points  moving  with  uniform  speed  v  start  at  the  same 
instant  in  the  same  direction  from,  the  point  of  contact  of  their 
paths.  The  one  moves  in  a  circle  of  radius  r,  the  other  in  a  tangent 
to  the  circle.    Find  their  relative  velocity  at  the  end  of  the  time  t. 

Ans.  2®  sin  k-  in  a  direction  inclined  to  the  tangent  at  an  angle  -in 1 

radians. 

•X'^i)  A  moves  N.E.  with  a  velocity  v,  and  B  moves  S.  15°  E.  with 
the  same  velocity.    Find  A's  velocity  relative  to  B. 

Ans.  V  t^S,  dxasHm^lStCS-      ^^""^ 

(13)  AxO'il^^y  train  runs  30  miles  per  hour  north.  Another 
running  is  miles  per  hour  appears  to  a  passenger  in  the  first  to  be 
running  at  25  mites  per  hour.  Find  the  direction  of  the  velocity  of 
the  latter.     ^^ 

Ans.  N.  56'  15  E.  or  N.  56'  15'  W. 

(14)  Two  candles  A  and  B,  each  1  ft.  long  and  requiring  4  and  6 
hours  respectively  to  burn  out,  stand  vertically  at  a  distance  of  I  ft. 
The  shadow  of  B  falls  on  a  vertical  wall  at  a  distance  of  10  ft.  from, 
B.    Find  the  speed  of  the  end  of  the  shadow. 

Ans.  8  inches  per  hour. 

(15)  A  ship  has  a  northeasterly  velocity  of  12  knots  per  hour 
Find  the  magnitude  of  her  velocity  {a)  in  an  easterly  direction; 
(6)  in  a  direction  15°  W.  of  N. 

Ans.  (a)  6  ^^2  ;  (b)  6  knots  per  hour. 

(16)  A  boat-crew  row  3i  miles  down  a  river  and  back  again  in  1 
hour  40  minutes.  If  the  river  has  a  current  of  2  miles  per  hour, 
find  the  rate  at  which  the  creiv  would  row  in  stilt  water. 

Ans.  5  miles  per  hour. 

v.,      (17)  A  steamer  goes  9.6  miles  per  hour  in  still  water.     How  long 
/V  will  it  take  to  run  10  miles  up  a  stream  and  return,  the  velocity  of 
r    the  stream  being  2  miles  an  hour  f 
Ans.  2  hours  11  minutes. 

(18)  A  steam  tug  travels  10  miles  an  hour  in  still  ivater,  but 
draws  a  barge  4  miles  an  hour.    It  has  to  take  the  barge  10  miles  tip 

,'ya  stream  ivhich  runs  1  mile  an  hour,  and  then  return  without  the 
Jv  barge.    How  long  will  it  take  for  the  job  f 
A         Ans.  4/5  hours. 

(19)  A  vessel  makes  two  runs  on  a  measured  mile,  one  with  the 
tide  in  a  minutes  and  one  against  the  tide  in  b  minutes.    Find  the 


CHAP.    IV.]  VELOCITY — EXAMPLES.  47 

velocity  of  the  vessel  through  the  water,  and  of  the  tide,  supposing 
both  uniform. 

Ans.  3Q- — — -  and  30  — --  miles  per  hour. 
^^    ab  ah 


(20)  A  point  receives  simultaneously  three  velocities,  60  ft.  per  sec. 
N.,  88  ft.  per  sec.  W.  30°  S.,  and  60  ft.  per  sec.  E.  30"  S.  Chive  the 
magnitude  and  direction  of  the  resultant  velocity. 

Aus^  feet  per  sec.  W.  30°  S. 


il)  A  ship  sailing  due  north  at  the  rate  of  8  miles  per  hour  is 
)^     carried  to  the  east  by  a  current  of  4  vniles  per  hour.    Find  the  ve- 
locity ivith  reference  to  the  land. 
Aus.  j3.9'i  miles  per  hour  N.  26°  34'  E. 

^-t§2)  A  ship  is  sailing  E.  22i°  S.  at  the  rate  of  10  miles  an  hour 
and  the  wind  seems  to  blow  from  the  N.  W.  unth  a  velocity  of  6  miles 
an  hour.    Find  the  actual  velocity  of  the  wind. 

Ans.  15.7  miles  an  hour  W.  30°  55'  N. 

(23)  A  point  moves  in  t  seconds  from  A  to  B,  the  positions  being 
given  by  the  co-ordinates  Xi ,  y^  and  Xt,  y^.  What  is  the  mean  ve- 
locity f 


V{Xi  —  Xiy-\-{y, pif^  making  an  angle  a  with  the  axis  of  x 

a  -yi 


Ans.  «  =  , 


given  by  tan  a  =  - 

X2  —  Xi 

^^  (24)  A  point  has  four  component  velocities  in  the  same  plane,  of 
12,  24,  36,  48  ft.  per  sec,  making  with  the  axis  of  X  the  angles  of 
16",  29°,  33°,  75°  respectively.     What  is  the  resultant  velocity  i 

Ans.  Vx  =  75.14;  Vy  =  80.915;  Vr  =  110.424;  angle  with  axis  of  x,  a  =  47" 
7'  10";  angle  with  axis  of  y,  b  =  42°  52'  50". 

C^  (25)  A  point  has  three  component  velocities  in  the  same  plane 
given  by  v,  =  40,  v-i  =  50,  Va  =  60  ft.  per  sec.,  maMng  with  the  axes 
of  X and  Y angles  given  by  ^i  =  60°,  /3,  obtuse;  /?»  =  30°,  as  obtuse; 
0-3  =  120°,  /Sa  obtuse.    Find  the  resultant  velocity. 

Ans.  We  have  /S,  =  150°,  a,  =  120°,  /?s  =  150°.  Hence  ij^;  =  -  35  ft.  per 
sec,  Vy  —  —  43.3  ft.  per  sec. ,  Vr  =  55.67  ft.  per  sec,  making  the  angles  with  X 
and  Fgiven  hy  a  =  128°  57'  17",  b  =  141°  2'  43 '. 

l^  (26)  A  point  has  the  component  velocities  u,  =  40,  Vi  =  50,  Vs  =  60 
ft.  per  sec.,  making  the  angles  with  X,  Y,  Z,  a,  —  60°,  /?i  =  100°,  xi 
obtuse  ;  a^  =  100°,  /?3  =  60°,  yt  acute  ;  as  =  120°,  /?3  =  100°,  yz  acute. 
Find  the  resultant  velocity. 

Ans.  We  find  the  angles  y  (page  12)  by  the  equation 
cos"  y  =  —  cos  {a  -\-  fi)  cos  (a  —  jS). 
Hence     y,  =  148°  2'  31"  7.    ya  =  31°  57'  28 '.3,    y^  =  31°  57'  28".3. 
Vx  =  —  18.6824  ft.  per  sec,  %  =  +  7.635  ft.  per  sec.  ,Vz  =  -\-  59.391  ft.  per  sec, 

Vr  =  62.73  ft.  per  sec, 
making  with  the  axes  of  X,  T,  Z,  angles  given  by 

a  =  107°  19'  36",      b  =  38°  0'  33",      c  =  18°  46'  42". 


pi 


Ak 


(f 


^M^ 


CHAPTEK  V. 


/IfV*- 


ACCELERATION. 


RESOLUTION    AND    COMPOSITION    OF 
ACCELERATIONS. 


ANALYTICAIi  DETERMINATION  OF  RESULTANT  FOR  CONCURRING  ACCELERA- 
TIONS. EQUATIONS  OP  MOTION.  THE  HODOGRAPH.  TANGENTIAL  AND 
NORMAL  ACCELERATION. 

Mean  Acceleration.— Let  P, ,  P2,  etc.,  Fig.  (a),  be  the  path  of  a 
moving  point  P,  and  let  the  corresponding  instantaneous  velocities 


Fig.  (a). 


Fig.  (6). 

in  the  time  i  is  a  = 


be  ^1,^2.  etc 

Each  velocity  is  tangent  to  the  path 
at  the  corresponding  point  and  is  equal 
in  magnitude  to  the  speed  at  that  point. 

If  t  is  the  number  of  units  of  time  in 
passing  from  Pi  to  P2 ,  the  mean  speed 

for  that  time  (page    15)  is  ^-^^-^i^ 

units  of  speed.  The  integral  change  of 
speed  in  the  time  t  is  t?2  —  Vi  (page  24), 
and  the  mean  rate  of  change  of  speed 


V-i  —  v> 
t 


(page  24). 


If  now  in  Fig.  (6)  we  draw  OQi  parallel  and  equal  in  magnitude 
to  Ui  and  OQi  parallel  and  equal  to  Vi ,  then  the  integral  change  of 
speed  is  represented  by  OQi  —  OQi  and  the  mean  rate  of  change  of 

speed  by '^^-Qg'. 

The  change  of  velocity,  however,  in  the  time  t  is  represented  in 
magnitude  and  direction  by  Qi  Q^ ,  and  this  we  call  the  integral  ac- 
celeration. 

The  mean  time-rate  of  change  of  velocity  is  given  in  magnitude 

by  ^y^\  and  in  direction  by  QiQi-    We  call  this  the  mean  linear 

acceleration.  The  term  acceleration  always  means  linear  accelera- 
tion unless  otherwise  specified. 

Mean  acceleration  is  then  time-rate  of  change  of  velocity,  whether 
that  change  takes  place  in  the  direction  of  motion  or  not. 

Instantaneous  Acceleration. — The  limiting  magnitude  and  direc- 
tion of  the  mean  acceleration,  when  the  interval  of  time  is  indefi- 
nitely small,  is  the  instantaneous  acceleration. 

The  limiting  direction  is  not  necessarily  tangent  to  the  path 

48 


€HAP.    v.]  ACCELERATION.  49 

except  in  the  case  of  rectilinear  motion,  and  the  limiting  magnitude 
is  not  the  rate  of  change  of  speed  in  the  path,  except  in  the  case  of 
rectilinear  motion. 

The  term  acceleration  always  signifies  instantaneous  accelera- 
tion unless  otherwise  specified.  It  is  the  limiting  time-rate  of 
change  of  velocity,  whether  that  change  take  place  in  the  direction 
of  the  motion  or  not. 

Acceleration  may  he  zero,  uniform  or  variable.  When  it  is  zero 
the  velocity  is  uniform,  and  we  have  uniform  speed  in  a  straight 
line. 

When  it  is  uniform,  it  has  the  same  magnitude  and  the  same 
unchanged  direction,  whatever  the  interval  of  time.  In  such  case 
the  acceleration  at  any  instant  is  equal  to  the  mean  acceleration  for 
any  interval  of  time.  If  its  direction  coincides  with  that  of  the 
velocity,  we  have  uniform  rate  of  change  of  speed  in  a  straight  line. 
If  it  does  not  so  coincide,  we  have  uniform  acceleration  and  motion 
in  a  curved  line. 

When  it  is  variable,  either  direction  or  magnitude  changes,  or 
both  change. 

Unit  of  Acceleration. — The  magnitude  of  the  imit  of  acceleration 
is  evidently  the  same  as  that  for  rate  of  change  of  speed,  viz.,  one 
unit  of  length-per-sec.  per  sec,  as  for  instance  one  foot-per-sec.  per 
sec.  We  denote  the  magnitude  of  the  acceleration  thus  measured 
by  the  letter/,  to  distinguish  it  from  rate  of  change  of  speed,  which 
we  have  denoted  by  a  (page  25). 

Line  Representative  of  Acceleration. — Since  acceleration  is  time- 
rate  of  change  of  velocity,  and  is  therefore,  like  velocity  and  dis- 
placement, a  vector  quantity,  it  can  be  represented  like  them  by  a 
straight  line,  whose  length  and  direction  give  the  magnitude  and 
direction  of  the  acceleration  (pages  34,  43).  Rate  of  change  of  speed 
is  given  by  stating  sign  and  magnitude  only.  It  is  a  scalar  quantity 
(page  25). 

Triangle  and  Polygon  of  Accelerations. —The  principles,  there- 
fore, of  page  35  hold  good  for  accelerations  also,  as  well  as  dis- 
placements. We  have  then  the  "  triangle  and  polygon  of  accelera- 
tions." 

Composition  and  Resolution  of  Accelerations. — For  the  same 
reason  we  can  covapine  and  resolve  accelerations  in  the  same  way 
as  displacements,  and  the  principles  of  pages 
35,  36  apply. 

Let  OB  and   OD  be  the  initial  and  final 
velocities  of  a  point  in  any  given  time  t. 

Then  BD  is  the  integral  acceleration  and 

TiD 

—  is  the  mean  acceleration,  or  the  instan- 
taneous acceleration  if  it  is  uniform.  If  not 
uniform,  ^^,  where  t  is  indefinitely  small,  gives  the  acceleration. 

Draw  OA  and  DC  at  right  angles  to  any  line  AO  through  B  in 
any  given  direction. 

Then  — ^  is  the  component  of  the  acceleration  in  this  direction. 

But  —  =  ^^~^,  and  AC  and  AB  are  the  components  of  the 

velocities  in  the  direction  AC. 

Hence  the  component  in  any  direction  of  an  acceleration  is  equal 
to  the  acceleration  of  the  component  velocities  in  that  direction. 


50  KINEMATICS — GENERAL   PRINCIPLES.  [CHAP.    Y, 

Sign  of  Components  of  Acceleration. — We  have  the  same  rule  for 
the  signs  of  the  horizontal  and  vertical  components  fx  and  fy  of  any 
acceleration/,  as  for  the  horizontal  and  vertical  components  of  Vx 
and  Vy  of  any  velocity  v  (page  44). 

If  the  direction  of  the  linear  representative  is  towards  the  right 
or  upwards,  fx  and  fy  are  positive ;  if  towards  the  left  or  doivn- 
ward,  fx  and  fy  are  negative. 

The  sign,  then,  applied  to  component  accelerations  indicates  di- 
rection of  action.  For  rectangular  co-ordi- 
nates (+)  signifies  in  the  direction  OX,  OY, 
OZ,  (— )  in  the  opposite  directions. 

If  polar  co-ordinates  are  used,  the  compo- 
nent acceleration  along  the  radius  vector  is 
positive  ( + )  when  it  acts  away  from  the  pole, 
and  negative  (— )  when  it  acts  towards  the 
pole. 

Evidently,  then,  we  measure  angles  in  the 
plane  XY  from  OX  around  towards  OY,  in 
the  plane  YZ  from  OY  around  towards  OZ, 
in  the  plane  ZX  from  OZ  around  toward  OX,  as  shown  by  the  ar- 
rows in  the  figure. 

Take,  for  instance,  the  case  of  a  particle  projected  vertically  away  from  the 
earth  with  the  initial  velocity  «,  and  attaining  the  final  velocity  v. 

As  long  as  the  particle  ascends,  the  direction  of  v  is  upwards,  and  v,  Vi  are 
both  positive.  The  acceleration  due  to  gravity  is  always  downwards,  and 
hence  is  negative. 

When  the  particle  is  descending  with  the  initial  velocity  Vi ,  then  both  Vi 
and  V  are  negative,  and  the  acceleration  is  negative  as  before. 

Analytical  Determination  of  the  Resultant  for  Concurring 
Accelerations. — When  the  line  representatives  meet  in  a  point,  they 
are  said  to  be  concurring.  We  have  then  the  same  expressions  for 
the  magnitude  and  direction  of  the  resultant  of  any  number  of  con- 
curring component  accelerations  as  for  displacements  (pages  37  and 
38).   We  have  only  to  substitute/ in  place  of  d. 

Equations  of  Motion  of  a  Point  under  Diflferent  Accelerations. — 
The  magnitude  of  the  acceleration  in  general  is  not  the  same  as 
rate  of  change  of  speed,  except  in  the  case  of  rectilinear  motion. 
We  have  therefore  denoted  it  by  /,  to  distinguish  it  from  rate  of 
change  of  speed,  which  we  have  denoted  by  a.  If,  then,  we  denote 
by  ft  the  magnitude  of  the  tangential  acceleration,  or  the  tangential 
component  of  /,  we  have  ft  =  a,  or  the  magnitude  of  the  tangential 
acceleration  is  equal  to  the  rate  of  change  of  speed. 

(a)  Acceleration  Zero. — We  have  in  this  ^ 
case  the  line  representative  QiQ'i  =  0,  and  ■„^^'']<P 
hence  the  line  representative  of  the  velocity           ^^"'^K^f^^^ 

does  not  change  either  in  direction  or  magni-      p-=i_ - >  ^q 

tude.  ''         ^^ 

We  have  then  rectilinear  motion  with  constant  velocity,  and  for 
any  interval  of  time 

S  —  Si  .^. 

^=-r-' ^'> 

where  s,  and  s  are  the  initial  and  final  distances  of  the  point  from 
any  fixed  point  in  the  line.  This  equation  is  general  if  we  pay 
attention  to  the  sign  of  displacement  and  velocity  (pages  37  and  44). 

(b)  When  the  Direction  of  the  Acceleration  coincides  with  the- 
Direction  of  the  Velocity. — In  this  case  the  line  representative  QiQt 


CHAP,    v.]  EQUATIONS   OF   MOTION.  51 

coincides  in  direction  with  PQi  —  Vi.  We  have  then  rectilinear 
motion  with  varying  velocity. 

If  /  is  uniform,  the  instantaneous  acceleration  is  equal  to  the 
mean  acceleration  for  any  interval  of  time. 

Hence,  if  u,  is  the  initial  and  v  the  final  velocity,  we  hsuveforf 
constant 

V  —  Vi 

f=—r- (2) 

This  value  of  /  is  general  when  we  pay  attention  to  the  sign  for 
velocity  and  acceleration  (pages  44  and  50). 
From  (2;  we  have 

v  =  vi  +ft (3) 

The  average  speed  is 

V+  Vi  1  ^^ 

~2~  ^^'  +  2^* (4) 

The  displacement  is 

s-s,='^^^t  =  v^t  +  ^ft' (5) 

Inserting  the  value  of  t  from  (2)  we  have 

v"  —  v,^ 
s-s,=      g^  ■ (6) 

Hence  v"  =  Vi""  +  2fis  —  s.) (7) 

[Tffis  variable,  we  have  from  (1),  in  Calculus  notation, 

ds 

^  =  ^' («) 

and  from  (3), 

f_dv_^^ 

and  from  (8), 

s-8,-  I  tdt (10) 


!  —  «1   =     /    « 


The  preceding  equations  can  be  deduced  from  these  as  on  page  28.] 
All  these  equations  are  precisely  similar  to  those  on  page  28,  except  that 
we  have  /  in  place  of  a. 

(c)  When  the  Acceleration  is  Constant  in  Magnitude  and  Direction, 
but  makes  an  Angle  (p  with  the  Initial  Velocity. — In  this  case  we 
have  motion  in  a  curve,  and  QiQ-i  =ft.    Hence 

v"  =  (vi  +  ft  cos  0)'  +  (ft  sin  (py  =  V,''  +  2vift  cos  <P  +  fH\    .     (11) 

The  tangent  of  the  angle  QiPQ^  is 

tan^.P^.  =  -^$^ (12) 

Vi+ft  cos  <t> 

The  displacement  in  the  time  t  is  given  by 

ip  =  Ut  +  ^/i^  cos  (p\\  (|/«'sin  0J';     .    .    .    .    (13) 

ahd  the  tangent  of  its  angle  of  inclination  to  Ui  is 

iff  sin  0 ,j^. 

Vit  +  Iff'  cos  0 


53 


KINEMATICS — GENERAL   PKINCIPLES. 


[chap.  V. 


The  Hodograph. — Let  a  point  P  moving  in  any  curve  have  the 
positions  Pi ,  I\,  Pz,  etc.,  and  let  the  magnitude  of  the  corresponding 

velocities  be  Vi,  zh,  fs,  etc.  These 
velocities  are  tangent  to  the  path  at 
P, ,  P2,  P3,  and  are  equal  in  magni- 
tude to  the  speed  at  these  points. 

If  from  a  point  O  we  draw  lines 
OQ,,  OQi,  OQ3,  etc.,  parallel  and 
equal  to  v,,  v^,  Vs ,  the  extremities  of 
these  lines  will  form  a  polygon. 

If,  however,  the  points  Pi,  P2,  P3 
are  indefinitely  near,  the  polygon  be- 
comes a  curve  ^1^2  ^3,  such  that  when 
the  point  P  describes  the  path  P1P2P3, 
we  can  conceive  another  point  Q  to  describe  the  curve  QiQiQs. 
This  curve  is  called  the  hodograph.*  The  point  O  is  called  the  pole 
of  the  hodograph.  The  points  Q,,  Q2,  Qs  are  called  the  points  cor- 
Tesponding  to  Pi,  P2,  Pa. 

Any  radius  vector,  as  OQ,,  OQ2,  in  the  hodograph,  represents  in 
magnitude  and  direction  the  velocity  at  the  corresponding  point 
Pi,  Ps,  etc.,  of  the  path.  The  magnitude  of  OQi,  OQn,  etc.,  is  the 
speed  y.,  v-^,  etc.,  at  P,  P,,  etc.  The  direction  of  O^,,  OQ^  is  the 
direction  of  v,,  Vi,  or  that  of  the  tangent  to  the  path  at  Pi ,  P2. 

If  t  is  the  interval  of  time  in  moving  from  P,  to  P2 ,  then  the 
chord  Q,  Q^  in  the  hodograph  gives  the  magnitude  and  direction  of 

the  integral  acceleration  for  that  time,  and ^^Vi  ^hich  is 

the  mean  velocity  in  the  hodograph,  gives  the  mean  acceleration  in 
the  path. 

When  the  time  is  indefinitely  small,  QiQi  becomes  tangent  to 

the  hodograph  and  ^'^'',  which  is  now  the  instantaneous  velocity 

in  the  hodograph,  gives  the  instantaneous  acceleration  in  the  path. 

The  tangent  to  the  hodograph  at  any  point,  therefore,  gives  the 
direction  of  the  instantaneous  acceleration  at  the  corresponding 
point  of  the  path.  The  instantaneous  speed  in  the  hodograph  gives 
the  magnitude  of  this  acceleration. 

Hence,  the  velocity  at  any  point  of  the  hodograph  is  the  accelera- 
tion at  the  corresponding  point  of  the  path. 

Tangential  and  Normal  Acceleration.— The  entire  resultant  ac- 
celeration /  at    any  point  of  the 
path  may  be  resolved  into  a  com- 
ponent   tangential  to  the  path  ft 
-and  a  component  normal   to   the 

path/n,  so  that/=  Vft"  +  fn'. 

The  magnitude  of  the  tangential 
-component  ft  is  the  rate  of  change 
of  speed  in  the  path.  Its  direction 
is  always  tangent  to  the  path,  or 
parallel  to  the  radius  vector  of  the 
hodograph  at  the  corresponding 
point  of  the  hodograph. 

In  order  to  find  the  normal  com- 
ponent fu ,  let  us  first  suppose  a  point  to  move  in  a  circle  with  con- 
stant  speed  v. 


*  Invented  by  Sir  W.  R.  Hamilton. 


CHAP,    v.]     TANGENTIAL   AND    NORMAL   ACCELERATION.  53 

Let  the  radius  of  the  circle  be  r,  and  take  any  two  points  Pi  and 
Pn  on  the  circle.  Then  the  velocity  at  Pi  is  v  tangent  at  Pi  or  per- 
pendicular to  r  at  that  point,  and  the  velocity  at  Ps  is  v  tangent  at 
Pa  or  perpendicular  to  r  at  A. 

Now  construct  the  hodograph  by  making  OQi  parallel  and  equal 
to  the  velocity  at  P  and  O^a  parallel  and  equal  to  the  velocity  at 
P2,etc. 

Evidently  the  hodograph  is  also  a  circle  of  radius  v,  and  the 
speed  in  the  hodogi'aph  is  also  constant,  since  the  point  P  moves 
with  constant  speed  and  makes  a  revolution  in  the  same  time  as  its 
corresponding  point  Q  in  the  hodograph.    Let  t  be  the  time  of  revo- 

27TV 

lution ;  then  -^  is  the  speed  of  Q  in  the  hodograph  or  the  accelera- 
tion of  P  in  the  path ;  and  since  this  speed  at  any  point  Q^  is  at  right 
angles  to  OQi  or  v,  it  is  normal  to  the  path  at  P  or  parallel  to  CP. 

We  have  then /n  =  ^-.    But  the  speed  in  the  path  is  v  — ——^ 

2tiv 
Hence  t  =  — ,  and  substituting,  we  have 

V 

We  can  obtain  the  same  result  as  follows  :  Since  the  point  P" 
moves  from  Pi  to  Pt  in  the  same  time  that  Q  moves  from  Qi  to  Q2 , 
and  the  angle  P1CP3  is  equal  to  the  angle  QiOQt,  we  have 

fn:  V-.-.V.r,     or    fn=  —  • 

r 

Since  we  have  supposed  v  constant  in  magnitude,  the  tangential 
acceleration  ft  is  zero,  and  therefore  fn  in  this  case  is  the  magnitude 
of  the  total  resultant  acceleration  /. 

A  normal  acceleration,  then,  has  no  effect  upon  the  speed,  but 
only  changes  the  direction  of  motion. 

Let  us  now  suppose  that  the  speed  v  is  not  constant,  the  point 
still  moving  in  a  circle.     Then  the  hodograph  will  not  be  a  circle. 

But  if  the  two  points  P  and  Pa  are  indefinitely  near,  so  that  the- 
arc  PiPi  is  indefinitely  small,  the  velocities  at  Pi  and  P  can  be 
taken  as  equal  still,  and  we  shall  still  have 

-  «        v" 

fn.  V-.-.V.r,     or     fn= 

r 

Again,  if  the  point  P  moves  in  any  curve  whatever,  a  circle  can 
always  be  described  whose  curvature  is  the  same  as  that  of  the 
curve  at  any  given  point.  The  radius  of  this  circle  is  the  radius  of 
curvature  p  at  the  point. 

In  this  case,  then,  we  should  have 

fn-.V-.-.V-.p,      or     fn——. 

Therefore,  in  general,  whatever  the  path  may  be  and  whatever 
the  speed  in  the  path — 

The  magnitude  at  any  instant  of  the  normal  component  of  the 
acceleration  is  equal  to  the  square  of  the  speed  at  that  instant 
divided  by  the  radius  of  curvature. 


54  KINEMATICS— GENERAL   rEINCIPLES,  [CHAP.  V. 


if^^h 


aM-*^         "^  EXAMPLES. 


# 
^ 


i 


(1)  A  point  moves  with  uniform  velocity  in  a  straight  line.  What 
is  the  hoaograph  f 
Ans.  A  point. 

^  (2)  A  point  moves  with  uniform  acceleration  either  in  a  straight 
line  or  a  curve.  What  is  the  hodograph  f  What  is  the  speed  in  the 
hodograph  f 

Ans.  A  straight  line.     Uniform. 

(3)  Show  that  the  direction  of  motion  of  any  point  B  on  the  cir- 
cumference of  a  circle  rolling  with  velocity  v  on  a  straight  line  is 
perpendicular  to  AB  at  any  instant,  when  A  is  the  point  of  contact 
of  the  circle  with  the  straight  line  at  the  instant  consider'ed. 

(4)  AB  is  a  diameter  of  a  circle  of  which  BC  is  a  chord.  When 
is  the  moment  about  A  of  a  velocity  represented  by  BC  the  greatest  f 

Ans^^-When  BC  and  AC  are  equal. 

U^)  A  point  is  moving  with  uniform  speed  of  a  mile  in  2  min. 
40  sec.  round  a  ring  of  100  ft.  radius.    Find  the  acceleration. 
An^^O.SQ  ft.-per-sec.  per  sec.  towards  the  centre. 

U^)  A  point  moves  in  a  horizontal  circle  ivith  uniform  speed  v, 
starting  from  the  north  point  and  moving  eastward.  Find  the 
integral  acceleration  when  it  has  moved  (a)  through  a  quadrant ; 
(b)  a  semicircle  ;  (c)  three  quadrants. 

Ans.  {a)  V  ^2^  SW. ;  (6)  2v,  W. ;  (c)  v  ^2,  NW. 

(7)  If  the  component  velocities  of  a  moving  point  are  represented 
by  the  sides  of  a  plane  polygon,  taken  the  same  way  round,  the 
algebraic  sum  of  their  moments  about  any  point  in  their  plane  is 

zero. 

Ans.  If  the  polygon  closes,  the  resultant  velocity  is  zero.  If  it  does  not 
close,  the  line  necessary  to  make  it  close  taken  the  other  way  round  is  the  re- 
sultant.    In  either  case  the  algebraic  sum  of  the  moments  is  zero. 

(8)  Show  that  the  hodograph  of  a  point  moving  with  uniform 
speed  in  a  circle  is  a  circle  in  which  the  corresponding  point  moves 
also  with  uniform  speed. 

(9)  Show  that  the  locus  of  the  extremity  of  a  straight  line  repre- 
senting either  of  the  tivo  equal  components  of  a  given  acceleration 
is  a  straight  line  perpendicular  to  the  straight  line  representing  the 
given  acceleration  and  through  its  middle  point. 

(10)  Let  the  velocity  of  a  point  moving  in  a  straight  line  vary  as 
the  square  root  of  the  product  of  its  distances  from  two  fixed  points 
in  the  line.  Show  thai  its  instantaneous  acceleration  varies  as  the 
mean  of  its  distances  from  the  fixed  points. 

Ans.  Let  «  be  one  distance  and  a  -f-  s  the  other.     Then  from  page  51,  Chap. 

V,  v  —  —  =  k  i^s{a  -[-  s),  where  k  is  a,  constant. 

_dv__  kads  +  2ksds  _  k^a  +  ^k^s _  k'^{a  -f  2«) 
^^"'~di~2dt  y^T^TsT  ~         2         ""         2        ' 

But  the  mean  of  the  distances  is  — —^ — — -  =  — - — . 


# 


^ 


^ 


i<t4r' 


CHAP,    v.]  ACCELEEATIOK — EXAMPLES.  55 

(11)  If  the  algebraic  sum  of  the  moments  of  the  component  ve- 
locities of  a  moving  point  about  any  two  points  P  and  Q  are  each 
zero,  shoto  that  the  algebraic  sum  of  their  moments  about  any  point 
in  the  line  PQ  ivill  he  zero. 

(12)  A  moving  point  P  has  tivo  component  velocities  one  of  which 
is  double  the  other.  The  moment  of  the  smaller  about  a  point  O  in 
the  plane  is  double  that  of  the  greater.  Find  the  magnitude  and  di- 
rection of  the  resultant  velocity. 

Ans.  If  a  is  the  smaller  component  and  a,  /?  are  the  inclinations  of  the 
greater  and  smaller  components  respectively  to  PO,  the  resultant  is 
a  i/b -\- 4  cos  {P -\- a),    and   it  is  inclined  to   PO  at  an   angle   whose  sin   is 

2  sin  a  -f-  sin  /H 
i/5+4cos(/i4-aj 

(13)  The  velocity  v  of  a  point  moving  in  a  straight  line  varies  as 
the  square  root  of  its  distance  sfrom  a  fixed  point  in  the  line.  What 
is  its  instantaneous  acceleration  f 

(is  — 

Ans.  We  have  «  =  — -  =  A;  Vs.  where  A  is  a  constant.     Hence 
at 

_dv  _     kds    _    kv    _  B*  _  A;' 

[4)  Tivo  raihcay  trains  move  in  directions  inclined  60°.  !Z%e 
one,  A,  is  increasing  its  speed  at  the  rate  of  ift.-per-min.  per  min. 
The  other,  B,  has  the  brakes  on  and  is  losing  speed  at  the  rate  of  8 
ft.-per-min.  per  min.     Find  the  relative  acceleration. 

Ans.  4  1/7  ft.  per  min.  per  min.,  inclined  to  the  direction  of  A  at  an  angle 
whose  sin  is  |/f  and  to  the  direction  of  B  at  an  angle  whose  sin  is  ^  y'f. 

•""(IH)  TTie  initial  and  final  velocities  of  a  moving  point  during  an 
interval  of  tivo  hours  are  8  miles  per  hour  E.  30°  N.  and  4  miles  per 
hour  N.     Find  the  integral  and  the  mean  acceleration. 

Ans.  4  4/8  miles  per  hour  W.,  2  V'3  miles-per-hour  per  hour  W. 

HIC^Ta  point  moves  in  a  circle  of  radius  8  inches  and  has  at  a 
given  position  a  speed  of  4  in.  per  sec,  which  is  changing  at  the 
rate  of  6  in.-per-sec.  per  sec.  Find  (a)  the  tangential  acceleration; 
(b)  the  normal  acceleration;  (c)  the  resultant  acceleration. 

Ans.  (a)  6  in.-per-sec.  per  sec;  (b)  2  in.-per-sec.  per  sec.;  (c)  2  ^10 in.-per- 
sec.  per  sec. 

(17)  Neivton  assumed  that  the  acceleration  of  gravity  varied  in- 
versely as  the  square  of  the  distance  from  the  earth's  centre.  He 
then  tested  this  assumption  by  applying  it  to  the  moon.  Assuming 
the  acceleration  at  tlie  earth's  surface  32.2  ft. -per -sec.  per  sec,  the 
radius  of  the  earth  4000  miles,  the  distance  betiveen  centres  of  earth 
and,  moon  240,000  miles,  and  the  speed  of  the  moon  in  its  orbit 
<i round  the  earth  3375  ft.  per  sec,  shoiv  that  Newton's  assumption  is 
in  accord  IV ith  fact. 

Ans     The  acceleration   of  the  moon's  centre  towards  the  earth  is  — ,  or 

r 

=  0.(X)89  ft.-per-sec.  per  sec. 


240000  X  5280 

But  according  to  Newton's  assumption,  if  g'  is  the  acceleration  at  the  dis- 

tance  of  the  moon,  ^  =  -^^^  =  3^.     Hence  g'  =  0.0089  ft.-per-sec 

per  sec. 


* 


* 


56  KINEMATICS— GENERAL  PRINCIPLES.  [CHAP.  V, 

(18)  Find  the  resultant  of  four  component  accelerations  repre- 
sented by  lines  drawn  from  any  point  P  within  a  parallelogram  to 
the  angular  points. 

Ans.  If  G  is  the  intersection  of  the  diagonals,  PC  represents  tlie  direction 
of  the  reetfTtant,  and  4PC  its  magnitude. 


(19)  A  ball  is  let  fall  in  an  elevator  which  is  rising  with  an  ac- 
celeration of  7.2  kilometers-per-min.  per  min.  The  acceleration  of 
the  ball  relative  to  the  earth  is  981  cm.-per-sec.  per  sec.  Find  its  ac- 
celeration relative  to  the  elevator. 

Ans.  1181  cm.-per-sec.  per  sec.  towards  the  floor. 

(20)  Assuming  the  mean  radius  of  the  earth  6370900  meters,  the 
speed  of  a  point  on  the  equator  465.1  m.  per  sec,  acceleration  of  a 
falling  body  9.81  m.-per-sec.  per  sec,  find  with  what  velocity  a  shot 
must  be  fired  at  the  eqtiator  with  either  a  westerly  or  easterly  direc- 
tion in  order  that,  if  unresisted,  it  shall  move  horizontally  round 
the  earth,  completing  its  circuit  in  1^  or  li  hours  respectively. 

Ans.  Westerly  velocity,  8370.7  meters  per  sec.  ;  easterly  velocity,  7440.5 
meters^ef  sec. 

''^1)  If  different  points  describe  different  circles  with  uniform 
speeds  and  with  accelerations  proportional  to  the  radii  of  their 
paths,  show  that  their  periodic  times  will  be  the  same. 

(22)  The  resultant  of  two  accelerations  a  and  a'  at  right  angles  to 
one  another  is  R.  If  a  is  increased  by  9  units  and  a'  by  5  units,  the 
magnitude  of  R  becomes  three  times  its  former  value,  and  its  direc- 
tion becomes  inclined  to  a  at  the  angle  of  its  former  inclination  to  a'. 
Find  a,  a'  and  R. 

Ans.  3,  4  and  5  units  respectively. 

(23)  If  a  tangent  be  drawn  at  any  point  of  a  conic  section,  the 
locus  of  the  foot  of  the  perpendicular  let  fall  from  a  focus  on  this 
tangent  is  a  circle  in  the  case  of  the  ellipse  and  hyperbola,  and  a 
straight  line  in  the  case  of  a  parabola.  Also  the  locus  of  the  foot  of 
a  perpendicular  from  the  vertex  of  a  parabola  on  a  straight  line 
drawn  through  the  focus  is  a  circle. 

Assuming  these  properties,  show  that  if  a  point  move  in  either  a 
circle,  ellipse,  hyperbola  or  parabola,  so  that  the  moment  of  its  ve- 
locity about  a  focus  is  constant,  the  hodograph  is  a  circle. 

(24)  Show  that  if  a  point  moves  in  an  ellipse  so  that  the  moment 
of  its  velocity  about  the  centre  is  constant,  the  hodograph  is  an 
ellipse.  [Note  that  the  area  of  the  parallelogram  formed  by  drawing 
tangents  to  an  ellipse  at  the  extremities  of  a  pair  of  conjugate 
diameters  is  constant] 

(25)  A  bullet  is  fired  in  a  direction  towards  a  second  bullet  which 
is  let  fall  at  the  same  instant.  Show  that  the  line  joining  them  icill 
move  parallel  to  itself  and  that  the  bullets  will  meet. 

(26)  Determine  whether  any  of  the  following  equations  are  pos- 
sible or  not : 

(1)  lOavst  +  Su's  =  ZgH* ; 

(2)  vH  —  4as  -I-  3a  =  0  ; 

(3)  6v  +  2g'asH'  =  Sa^'st*. 

Ans.  The  first  gives  us  !:— j^  in  each  term  and  is  therefore  possible.  The 
second  gives  us  ^tj—,  ^—^  and  j-iy?rj,  or  each  term  refers  to  different  kinds  of 


CHAP,   v.]  ACCELERATIOIf — EXAMPLES.  57 

quantities,  and  the  equation  is  nonsense  on  its  face.  The  third  gives  us  ~^, 
[i]"  and  [LY,  and  is  also  nonsense. 

(27)  A  point  moves  in  a  straight  line  so  that  the  number  of  units 

of  distance  sfrom  the  origin  at  the  end  of  any  number  of  seconds  t 

5         8  5 

is  given  by  s  =  2  +  -t  -] — P  +  -<*.    Find  (a)  the  number  of  units  of 

distance  from  the  origin  at  the  start ;  (b)  the  velocity  v  at  any  in- 
stant ;  (c)  the  acceleration  a  at  any  instant ;  (d)  the  velocity  at  the 
start ;  (e)  the  acceleration  at  the  start. 

5       3         15  3       15 

Ans.    (a)  2  units  of  distance;    {b)   v  =  -^-\- -xt-^  -^f^  ;  (c)  «  =  „  +   j^J 

5  3 

(d)  -  units  of  distance  per  sec. ;  {e)  ^  units  of  distance-per-sec.  per  sec. 

(28)  A  point  moves  in  a  straight  line  so  that  the  number  of  units 
of  acceleration  a  at  the  end  of  any  number  of  seconds  t  is  given  by 

a  =  7  —  —t  -¥  2<^    If  Vi  is  the  number  of  units  of  velocity  at  the 

o 
start,  and  Si  the  number  of  units  of  distance  from  the  origin  at  the 
start,  find  the  velocity  anid  the  distance  from,  the  origin  at  any  in- 
stant. 

Ans.    v  =  Vi-\-lt-^t''-\-  y? ; 

(29)  A  point  moves  in  a  straight  line  so  that  the  number  of  units 

of  velocity  v  at  the  end  of  any  nuniber  of  seconds  t  is  given  by 

3        5 
V  =  5  —  -t  +  -t^.    Find  the  acceleration  a  and  the  distance  s,  if  Si  is 
2        6 

the  initial  distance. 

Ans.  a  =  _  -  ^-  -«; 

(30)  A  point  has  three  component  accelerations  in  the  same  plane 
given  by  f\  =  40,  /a  =  50,  fa  =  60ft.-per-sec.  per  sec,  making  with  the 
axes  of  X  and  Y  angles  given  by  oci  =  60°,  y3i  obtuse;  02  =  30%  «* 
obtuse;  as  =  120°,  /?3  obtuse.    Find  the  resultant  acceleration. 

Ans.  We  have  /?i  =  150°,    a^  =  120°,    ^3  =  150°.     Hence 

fx  =  —  35  ft.-per-§ec.  per  sec. ;    /»/  =  —  43.3  ft.-per-sec.  per  sec,   and 
fr  —  55.67  ft.-per-sec.  per  sec, 

making  the  angles  with  X  and  Y  given  by 

a  =  128°  57'  17",     b  =  141°  2'  43". 

(31)  A  point  has  three  component  accelerations,  f,  =  40,  fi  =  50, 
fs  =  QO  ft.-per-sec.  per  sec,  making  with  the  axes  of  X,  Y,  Z  angles 
given  by  a,  =  60°,  ^,  =  100',  y,  obtuse  ;  a-i  =  100%  (ii  =  60°,  y^  acute; 
as  —  120°,  hi  =  100°,  yi  acute.     Find  the  resultant  acceleration. 

Ans.  We  find  the  angles  y  (page  12)  by  the  equation 

cos"  y  =  —  cos  (a  -\-  (i)  cos  (a  —  /8). 


58  KINEMATICS — GENERAL   PRINCIPLES.  [CHAP.  V, 

Hence     yx  =  148°  2'  31".7,    y^  =  31°  57'  28".3,     y^  =  31°  57'  28".3 ; 
fx=-  —  18.6824  ft.-per-sec.  per  sec,    fy  =  -\-  7.635  ft.-per-sec.  per  sec., 
fz  =-\-  59.391  ft.-per-sec.  per  sec,     fr  =  62.73  ft.-per-sec.  per  sec, 

making  with  the  axes  of  X,  Y,  Z  angles  given  by 

a  =    07°  19'  36  '.     6  =  83°  0'  33",     c  =  18°  46'  42". 

(32)  Investigate  the  motion  of  a  point  whose  initial  velocity  in  a 
horizontal  direction  is  0,  and  in  a  vertical  direction  —  32  ft.  per  sec. 
The  horizontal  acceleration  is  fx=  +  16  ft.-per-sec.  per  sec.,  and  the 
vertical  acceleration  fy  =  +  At  ft.-per-sec.  per  sec. 

Ans.  The  resultant  acceleration  is  (page  51) 


ajid  it  makes  an  angle  A  with  the  horizontal  whose  cosine  is 
cos  A  = 


4Uid  an  angle  pi  with  the  vertical  whose  cosine  is 

cos  yU 


4/(16)" +  (4<)* 
The  horizontal  velocity  at  the  end  of  any  time  is 

vx  =  16«. 
The  vertical  velocity  at  the  end  of  any  time  is 

Cj,  =  _  32  +  2t\ 
The  resultant  velocity  is 


V  =  Vil^ty  +  (2t^  -  32)2  =  3«»  4. 3S^ 

imd  it  makes  an  angle  a  with  the  horizontal  whose  cosine  is 

8t 
««««  =  , -^+16' 

and  an  angle  fi  with  the  vertical  whose  cosine  is 

^°^^  =  ^-l6- 
The  distance  s  described  in  any  time  is 

«=  |<3+32«. 


The  tangential  acceleration  is 
ft  =  a  =fx  cos  a  +/vCOJ 

The  normal  acceleration  is  (page  52) 


fi  =  a=f.  cos  a  +/„cos  /?  =  16  X  ^T^  +  4<  X  ^q:^  =  4t 


"The  radius  of  curvature  is  (page  53) 

g»      (2<'4-32)* 

^~fn~  16  • 

The  horizontal  distance  described  is 
X  =  8t\ 


CHAP,    v.]  ACCELERATION — EXAMPLES.  59 


The  vertical  distance  described  is 


y  =  |<^  -  S2t. 
Eliminating  t,  we  have  for  the  equation  of  the  path 

(33)  Investigate  the  motion  of  a  point  whose  initial  velocity  in 
the  direction  of  the  axis  of  x  is  +  2  ft.  per  sec.;  in  the  direction  of 
the  axis  of  y,  0]  in  the  direction  of  the  axis  of  z,  +4  ft.  per  sec. 
The  acceleration  in  the  direction  of  the  axis  of  xisfx  =  0;  in  the  di- 
rection of  the  axis  of  y,fy=  +  Sft.-per-sec.  per  sec.;  in  the  direction 
of  the  axis  of  z,fz=  +  5  ft.-per-sec.  per  sec. 

Ans.  The  resultant  acceleration  is  (page  51) 


/=  -^/0''  +  3''  +  5''=   i/34, 
which  makes  an  angle  A.  with  the  axis  of  x  whose  cosine  is 

cos  A  =  — -=  =  0 ; 
i/34 

with  tho  axis  of  y  an  angle  /i  whose  cosine  is 

8 


cos  JJ 


with  the  axis  of  g  an  angle  v  whose  cosine  is 

5 

cos  V  =   r=r. 

i/34 
The  velocities  in  the  direction  of  x,  y  and  z  are 

Vx  =  3;    ?)y  =  3<;    ««  =  4  +  Sit 
The  resultant  velocity  is 


«  =  |/2!>  +  {^tf  +  (4  +  Uf  =  V'34«»  +  40«  +  20, 
-which  makes  angles  a,  /3  and  y  with  the  axes  of  x,  y  and  z  given  by 

2  ,,  m 

cos  p 


4/34««  +  40«  +  20  V^f + 40«  +  30 

4  +  5« 
cos  y 


>^34i»  +  40«  +  20 
The  distances  described  in  tho  direction  of  x,  y  and  z  are 

x^2t;     y  =  |«';    8  =  4«  +  |<«. 

If  we  eliminate  t,  we  have 

y=-^; 

2  =  2a;  +  -5<B». 

o 

These  are  the  equations  of  the  projection  of  the  path  upon  the  planes  of  vy 
and  OSS. 


(0 


^^y^ 


r^ 


CHAPTEE  VL 


MOMENT  OF  DISPLACEMENT,  VELOCETY,  ACCELERATION. 
LINE  REPRESENTATIVE. 

KESOLUTION  AND  COMPOSITION  OF  MOMENTS  OF  DISPLACEMENT,  VELOCITY, 
ACCELERATION.  ANALYTICAL  DETERMINATION  OP  RESULTANT  VELOCITY 
AND  MOMENT  FOR  CONCURRING  VELOCITIES,  ACCELERATIONS  AND  DIS- 
PLACEMENTS. 

Moment  of  Displacement,  Velocity  or  Acceleration  about  a  Given 
Point. — The  product  of  the  magnitude  of  a  displacement,  velocity 
or  acceleration  by  the  magnitude  of  the  perpendicular  let  fall  from 
any  given  point  upon  its  direction  gives  the  magnitude  of  the 
moment  of  the  displacement,  velocity  or  acceleration  with  reference 
to  that  point. 

The  perpendicular  is  called  the  lever  arm,  and  the  point  is  the 
centre  of  moments. 

Thus  if  AB  is  the  line  representative  of  any  displacement,  ve- 
locity or    acceleration,    and  the  perpendicular 
?  from  any  point  O  is  p,  the  moment  relative  to  O 

/  is  ±  AB  X  p.     If  AB  is  a  displacement  d,  we 

/  have  ±  dp.     If  it  is  a  velocity  v,  we  have  ±  vp. 

/     ^^        \      If  it  is  an  acceleration/,  we  have  ±  fp. 
l^,"'^^  /  1+         The  (+)  sign  indicates  that  the  radius  vector 

o*^ 7A  I      moves    around  O  counter-clockwise   as  in  the 

^\y  figure.    The  (— )   sign  denotes  clockwise  rota- 

tion. 

If  we  draw  a  line  00'  through  O  parallel  to  AB,  it  is  evident 
that  the  moment  of  AB  is  the  same  relative  to  any  point  in  this  line. 

Hence,  so  far  as  the  moment  is  concerned,  a  displacement, 
velocity  or  acceleration  may  be  considered  as  laid  off  from  any 
point  in  its  line  of  direction. 

Moment  of  Displacement,  Velocity  or  Acceleration  about  a  Given 
Axis. — Let  OF  be  a  given  axis  and  y 

AB  any  displacement,  velocity  or  ac- 
celeration. 

Take  any  plane  XZ  perpendicular 
to  the  axis  OY,  intersecting  this  axis 
in  O.  Project  AB  upon  this  plane  in 
A!B'.  Now  AB  can  be  resolved  at  A 
into  a  component  parallel  to  AB'  and 
a  component  perpendicular  to  A!B' 
and  therefore  parallel  to  the  axis  OY. 

The  component  parallel  to  OY 
causes    no    rotation    about    Y.    The  . 

moment  of  AB  about  the  axis  is  then  the  moment  of  A!B'  aboutthe 
intersection  O  of  the  plane  XZ  with  the  axis. 


CHAP.  VI.]  MOMENT   OF    DISPLACEMENT.  6^ 

Hence  the  moment  of  a  displacement,  velocity  or  acceleration 
about  a  given  axis  is  the  moment  of  the  component  in  a  plane  per- 
pendicular to  the  axis  about  the  point  of  intersection  of  that  plane 
with  the  axis. 

Here  again,  so  far  as  moment  about  any  axis  is  concerned  a 
displacement,  velocity  or  acceleration  may  be  taken  as  laid  off  from 
any  point  in  its  line  of  direction. 

Moment  of  Displacement.— If  AB  =  d  is  a  displacement,  we  have 
then  for  its  moment  about  any  point  O, 
il/  =  ±  dp,  and  dp  is  twice  the  area  of 
the  triangle  AOB. 

Hence  the  moment  of  a  displacement 
about  any  point  is  twice  the  area  swept 
through  by  tlie  radius  vector  from  that 
point. 

The  unit  of  moment  of  a  displace- 
ment is  then  the  square  of  the  unit  of         p\ 
length,  as  one  square  foot.  \      /' 

If  we  draw  00'  parallel  to  AB  and  V' 

erect  a  perpendicular  DO'  at  the  middle 

point  of  AB,  then  DO'  =p  and  the  moment  about  O  is  equal  to  the 
moment  about  any  point  in  the  line  0(7.  If  O'A  —  O'B  =  r,  and 
the  angle  AO'B  =  9,  then  we  have  for  the  displacement  d 

d  =  2r  sin  ^. 

Moment  of  Velocity — Acceleration— If  AB  =  u  is  a  velocity,  we 
have  for  its  moment  about  any  point  O,  M=  ±  vp. 

Hence  the  moment  of  a  velocity  about  any  point  is  tvnce  the 
areal  velocity  of  the  radius  vector. 

The  unit  of  moment  of  a  velocity  is  then  the  square  of  the  unit 
of  length  per  unit  of  time,  as  one  square  foot  per  second. 

If  AB  =  /■  is  an  acceleration,  we  have  for  its  moment  about  any 
point  O,  M—  ±fp. 

Hence  the  moment  of  acceleration  about  any  point  is  twice  the 
areal  acceleration  of  the  radius  vector. 

The  unit  of  moment  of  an  acceleration  is  then  the  square  of  the 
unit  of  length  per  unit  of  time  squared,  or  one  square  ft.-per-sec. 
per  sec. 

Line  Representative  of  Moment  of  Displacement — Velocity — Ac- 
celeration.— Since  moment  of  displacement,  velocity,  acceleration, 
has  thus  both  magnitude  and  direction,  it  is  a  vector  quantity  like 
displacement,  velocity,  acceleration  themselves,  and  like  them  can 
be  completely  represented  by  a  straight  line. 

Thus  the  length  of  a  line  represents  the  magnitude  of  the 
moment  dp,  vp  or  fp.  This  line  is  always  taken  perpendicular  to 
the  plane  of  rotation  of  the  radius  vector,  which  is  therefore  known 
when  the  direction  of  the  line  representative  is  known.  Finally  we 
denote  the  direction  of  rotation  in  the  plane  by  placing  an  arrow- 
head on  the  line,  so  that  wJien  ive  look  along  the  line  in  the  direction 
of  this  arrow,  the  rotation  is  always  seen  clockwise. 

s  Thus  the  line  AB  denotes  by  its  length  the  magni- 
tvide  of  the  moment  dp,  vp  or  fp.  The  plane  of  rotation 
of  the  radius  vector  is  at  right  angles  to  this  line,  which 
is  therefore  coincident  with  the  axis  of  rotation.  The 
direction  of  rotation  is  clockwise  in  this  plane  when 
^  we  look  from  A  to  B. 


62  i£INEMATICS— GENEEAL   PRIKCIPLE8.  [CHAP.  VI.. 

Resolution  and  Composition  of  Moments. — The  principles,  there- 
fore, of  pages  35  and  36  hold  good  for  moments  of  displacements, 
velocities  and  accelerations,  as  well  as  for  displacements,  velocities 
and  accelerations  themselves.  We  have  then  the  triangle  and 
polygon  of  moments. 

Sign  of  Components  of  Moments. — The  signs  of  the  line  repre- 
sentatives of  the  components  along  the  axes  of  X,  F,  Z  of  a  moment 
of  displacement,  velocity  or  acceleration 
foliow  the  same  rule  as  for  components  of 
displacement,  velocity  or  acceleration  (pages 
36,  44,  50). 

Hence  components  in  the  directions  OX, 
OY,  OZ  are  positive  (  +  ),  in  the  opposite 
direction  negative  (— ).  If  then  we  look 
along  the  line  representatives  of  the  compo- 
nents towards  the  origin  O,  the  rotation  is 
always  seen  counter-clockwise.  Therefore 
rotation  from  X  towards  F,  F  towards  Z,  Z 
towards  JTis  positive,  in  the  opposite  direc- 
tions negative. 

For  polar  co-ordinates,  directions  away  from  the.  pole  are  posi- 
tive, towards  the  pole  negative.* 

The  algebraic  sum  of  the  moments  of  any  number  of  component 
displacements,  velocities  or  accelerations,  about  any  point  in  their 
plane,  or  about  any  axis,  is  equal  to  the  moment  of  the  resultant 
displacement,  velocity  or  acceleration  about  that  point  or  axis. 

Let  AB,  AC  represent 
two  component  displace- 
ments, velocities  or  acceler- 
ations of  a  point  A.  Then 
the  resultant  is  AR.  Let  O 
be  any  point  in  the  plane  of 
the  components  either  out- 
side or  inside  the  angle  be- 
tween the  resultant  and  either  component. 
Then  in  the  first  case 

area  OAR  =  area  OAB  +  area  BAR  —  area  ROB, 
and  in  the  second  case 

area  OAR  =  area  OAB  +  area  ROB  —  area  BAR. 
In  both  cases 

area  BAR  =  area  ROB  +  area  OAC, 

since  all  three  triangles  have  equal  base  BR,  and  the  altitude  o!: 
BAR  is  the  sum  of  the  altitudes  of  ROB  and  OAC. 
We  have  then  in  the  first  case 

area  OAR  =  area  OAB  +  area  OAC, 

and  in  the  second  case 

area  OAR  =  area  OAB  —  area  OAC. 

*  Evidently,  then,  we  measure  angles  in  the  plane  X I^  from  OX  around 
towards  OF  ;  in  the  plane  YZ  from  OT  around  toMip,rds  OZ ;  in  the  plane  ZX 
from  OZ  around  towards  OX,  as  shown  by  the  arrows  in  the  figure. 


CHAP.  VI.]    RESOLUTION   AND   COMPOSITION   OF   MOMENTS. 


63. 


But  twice  the  area  OAR  is  the  moment  of  the  resultant  AH,  and 
twice  the  areas  OAB  and  OAC  are  the  moments  of  the  componenta 
AB  and  AC  about  O.  Hence  the  moment  of  the  resultant  is  equal 
to  the  algebraic  sum  of  the  moments  of  the  components. 

If  we  have  a  third  displacement,  velocity  or  acceleration  at  A, 
the  resultant  of  this  and  AR  would  be  the  resultant  for  all  three. 
Hence  the  principle  holds  for  any  number  of  components. 

Again,  let  AB,  BC,  CD  represent 
the  components  of  a  point  A.  Then 
the  resultant  is  AD.  Let  OY  be  an 
axis  and  XZ  a  plane  perpendicular  to 
the  axis.  Let  ab,  be,  cd  be  the  projec- 
tions on  this  plane  of  the  component 
velocities. 

We  have  just  proved  that  the  mo- 
ment of  ad  about  O  is  the  algebraic 
sum  of  the  moments  of  the  compo- 
nents ab,  be,  cd. 

But  the  moment  of  each  of  these  " 
about  O  is  the  moment  of  AB,  BC,  CD  about  the  axis  (page  60). 
Hence  the  moment  of  the  resultant  AD  about  the  axis  is  equal  to 
the  algebraic  sum  of  the  moments  of  the  components  AB,  BC,  CD. 
The  moment  of  acceleration  of  a  moving  point  relative  to  any 
fixed  point  in  the  plane  of  its  motion  is  equal  to  the  time-rate 
of  change  of  the  moment  of  its  velocity  about  the  same  point. 

Let  AB  =  t;i  be  the  instantaneous  velocity  of 
a  point  A  and  f  its  instantaneous  acceleration. 
Then  in  any  indefinitely  small  time  dt  the 
change  of  velocity  is  BC  =fdt,  and  the  resultant 
velocity  is  .AC  =  -y  in  the  plane  of  y,  and  /.  Take 
a  point  O  in  the  same  plane  and  drop  the  per- 
pendiculars li ,  I  and  p  upon  the  directions  of 
Vi ,  V  and/. 

Then,  since  the  moment  of  the  resultant  is  equal  to  the  algebraic 
sum  of  the  moments  of  the  components,  we  have 


vl  =  Vili  +  fdt .  p,  or  fp 


Vl  —  Vili 

di      • 


If  the  path  is  a  circle  of  radius  r,  then  l  =  li  =r,  and  we  have 
relative  to  the  centre 

fp=ftr. 

We  obtain  the  same  result  as  follows  : 

Resolve  the  acceleration  /  into  components  ft 

tangent  to  the  circle  and  fn  normal.     The  latter 

component    passes    through  the  centre,   and  its 

moment  is  zero.    We  have  then  the  moment  of  / 

Y  equal  to  the  moment 

Vy  Vr 


Pi    *^  "   \ 


of   the   other 
ponent,  or 

fp=ftr. 
General  Analytical  Determination 
of  Resultant  Velocity  and  Moment 
for  Any  Number  of  Concurring  Com- 
ponent Velocities  in  the  Same  Plane. 
—Let  the  point  P  be  given  by  the 
co-ordinates  x,  y.    Let  the  component  velocities  of  P  be  tJi ,  Ua ,  v» » 


64  KINEMATICS — GENERAL   PRINCIPLES.  [CHAP.    VI. 

«tc.,  all  in  the  same  plane  XY  and  making  with  the  axes  of  X  and 
Fthe  angles  cxi ,  /J,  ;  a., ,  /ia ;  tf^s ,  /is ,  etc. 

Let  V  be  the  resultant  velocity  making  the  angles  a  and  &  with 
the  axes  of  X  and  i'  respectively. 

For  the  component  velocities  parallel  to  X  and  Y  we  have 

Vx  =  V,  cos  a,  +  V2  cos  cr^  +  Vi  COS  CVa  +  .   .  .  =  ^UCOS  «;   \ 
Vy  =  Vi  COS  /^j  +  Vi  COS  /?2  +  ?'3  COS  pa  +  .  .  .  =  2v  COS  /i.   f 

In  these  summations  components  towards  the  right  or  upwards 
are  positive,  towards  the  left  or  downwards  negative. 
The  resultant  velocity  is  then 


Vr  =  Vvx'  +  Vy\ (2) 

making  with  the  axes  of  X  and  Y  angles  a  and  6  given  by 

cos  a  =  — ,    cos  b  =  ~: (3) 

Vr  Vr 

The  moment  of  the  resultant  velocity  with  reference  to  O  is  the 
algebraic  sum  of  the  moments  of  the  component  velocities  Vx  and 
Vy.  If  p  is  the  lever  arm  of  v,. ,  we  have,  paying  regard  to  the  sign 
for  du'ection  of  rotation,  for  the  moment  Mz  about  the  axis  OZ,  or 
the  moment  in  the  plane  XY, 

Mz  =  VrP=VyX  —  Vxy (4) 

Hence  the  lever-arm  is 

Mz  _, 

p  =  — (5) 

^         Vr 

In  these  equations  Vx,  Vi,,  x  and  y  are  positive  in  the  directions 
OX,  OF  and  negative  in  the  opposite  directions.  With  these  con- 
ventions the  equations  are  general. 

If  Mz  comes  out  positive,  the  direction  of  rotation  about  O  is 
counter-clockwise;  if  negative,  the  direction  of  rotation  is  clock- 
wise as  shown  in  the  figure. 

In  the  first  case  the  line  representative  is  positive  and  therefore 
passes  through  O  in  the  direction  OZ.  In  the  second  case  it  has 
the  contrary  direction.  In  both  cases,  if  we  look  along  the  line  rep- 
resentative towards  the  origin,  the  direction  of  rotation  will  be 
seen  as  counter-clockwise. 

Since  Vr  may  be  considered,  so  far  as  the  moment  at  O  is  con- 
cerned, as  acting  at  any  point  in  its  line  of  direction  (page  60), 
let  us  take  it  acting  at  E,  the  intersection  of  the  line  of  direction 
of  V  with   the  axis  of    Y.    Then  we  have  for  the  distance  OE, 

Mz 
Vx  X  OE  =  —  Mz,  or  OE  = .    The  tangent  of  the  angle  which 

Vti 

Vr  makes  with  the  axis  of  X  is  -— .    Hence  the  equation  of  the  line 

Vx 

of  direction  of  the  resultant  velocity  v  is 

Vy  Mz  ,„. 

y  =  -x  —  — (6) 

Vx  Vx 

If  in  this  equation  we  make  x  =  0,  we  find  the  ordinate  of  the 
point  in  which  the  direction  of  the  resultant  velocity  Ur  intersects 
the  axis  of  Y,  viz., 

OE  =  y'=-'^ (7) 

Vx 


<:HAP.  VI.]  MOMENT   OF   VELOCITY.  65 

If  we  make  y  =  0,  we  find  the  abscissa  of  the  point  in  which  the 
-dii-ection  of  the  resultant  velocity  v,-  intersects  the  axis  of  X,  viz., 

X  =  — (8) 

General  Analytic  Determination  of  Resultant  Velocity  and  Mo- 
ment for  Concurring  Component  Velocities  not  in  the  Same  Plane. — 

Let  the  point  P  be  given  by 
the  space  co-ordinates  x,   y,  z,  T  x  . .. 

and  let  the  component  velocities 
of  P  be  y. ,  ^2,  fa,  etc.,  making 
with  the  axes  of  X,  Y,  and  Zthe      *-m 
angles  ^r, ,  /i, ,  :ki  ;  as ,  /ij ,  /'a  ;  as, 
/-^3,  ya,  etc. 

Let  i?rbethe  resultant  velocity 
making  the  angles  a,  6,  c  with 
the  axes. 

We  have  then  for  the  compo- 
nent velocities  parallel  to  X,  Y.  z  ^^^-'T^ 
and  Z                                                                                -^^My 

Vx  =  Vi  COS  Oil  +  Vi  cos  aa  +  Vi  COS  as  =  ^V  COS  oc ;  ^ 

Vy  =  V,  COS  /3,   +  V'2  COS  A  +  Us  COS  /?3  =  2VC08/3;  (     .      .      (1) 

Vz  =  u,  cos^i  +  ??2Cosr2  +  Vs COST'S  =  ^vcosy.  ) 

In  these  summations  components  in  the  directions  OX,  OY,  OZ 
are  positive,  in  the  opposite  directions  negative. 
The  resultant  velocity  is  then 


"/ 


Vr  =   I  Vx"  +  Vj,"  +  t?2*, (2) 

making  with  the  axes  of  X,  Fand  Z  angles  a,  6  and  c  given  by 

cosa=-,    cos  6=  J'-,    cosc=-- (3) 

Vr  Vr  Vr 

The  moment  of  the  resultant  velocity  Vr  with  reference  to  O  is 
the  algebraic  sum  of  the  moments  of  the  component  velocities  Vx , 
Vy  and  Vz . 

We  take  the  positive  direction  of  rotation  in  each  of  the  co- 
ordinate planes  in  the  direction  indicated  by  the  arrows  in  the 
figure.     Thus, 

rotation  about  Z  from  Xto  Y') 

"  "     X     "     F  "  Z  >•  are  positive ; 

"     Y      "    Z '' X) 

in  the  contrary  directions,  negative. 

We  have  then  for  the  moments  about  the  axes 

moment  about  Z parallel  to  plane  XY,  Mz  =  VyX  —  Vxy;  ) 

"     X       "        "      "      YZ  Mx  =  vzy-Vyz-A.  .    .    (4) 
"     Y        "        "      "     ZX,  M„  =  VxZ  -  VmX.  ) 

In  these  equations  Vx ,  Vy ,  Vz  and  x,  y,  z  are  positive  in  the  direc- 
tions OX,  OY,  OZ,  and  negative  in  the  opposite  directions.  With 
these  conventions  the  equations  are  general;  and  if  Mz  Mx,  My 
come  out  positive,  we  have  rotation  in  each  plane  counter-clock- 
wise as  indicated  by  the  figure ;  if  negative,  clockwise. 

In  the  first  case  the  line  representatives  pass  through  O  and 
have  the  directions  OZ,  OX,  OY.  In  the  second  case  they  have 
opposite  directions  through  O.    In  any  case  the  direction  of  rota- 


66 


KINEMATICS — GENERAL   PRINCIPLES.  [CHAP.    VI. 


tion  is  always  counter-clockwise  when  we  look  along  the  line  repre- 
sentative towards  O. 

The  equations  of  the  projections  of  the  line  of  direction  of  the 
resultant  velocity  v,-  upon  the  co-ordinate  planes  are  found  as  in 
the  preceding  Article,  since  each  may  be  considered  as  acting  at 
any  point  in  its  line  of  direction  : 

Vy  Mz 

on  plane  XY,  y  =  -^-x : 

Vx  Vx 

"      "      YZ,z  =  -^y-^: 

Vy  Vy 

My_ 

Vz  ' 


ZX,   X  =  ^z 

Vz 


(5) 


If  in  these  equations  we  make  2;  =  0,  we  find  the  co-ordinates  of 
the  point  in  which  the  direction  of  the  resultant  velocity  Vr  pierces, 
the  plane  XY,  viz., 


0!'  =  - 


Vz' 


^  Vz 


(6) 


If  we  make  x  =  0,  we  have  for  the  co-ordinates  of  the  point 
where  it  pierces  the  plane  YZ, 

My 


z  = 


y  = 

Vx  Vx 


(7) 


If  we  make  2/  =  0,  we  have  for  the  co-ordinates   of  the  point 
where  it  pierces  the  plane  ZX, 

z'  =  -^,    x'  =  ^ (8) 

Vy  Vy 

Combining  the  line  representatives  of  the  moments  given  by 
(4)  we  have 

Mr  =  VrP  =  VMx'  +  My'  +  Mz' (9> 

Hence 

Mr 


P  = 


Vr 


(io> 


The  line  representative  for  the  resultant  moment  Mr  passes 
through  O  and  makes  the  angles  d,  e,  f  with  the  axes  of  X,  F,  Z 
given  by 


cos  d  = 


Mx 

Mr' 


COS  e  = 


_My_ 
Mr' 


cos/ 


Mz 

Mr 


(11) 


Looking  along  this  line  representative  towards  O,  the  direction 
of  rotation  is  always  counter-clockwise. 

The  projections  of  this  line  representative  upon  the  co-ordinate 
planes  make  angles  with  the  axes  given  as  follows: 


M 
projection  on  XY  tangent  of  angle  with  X=  j^- 

Mx 


YZ 
ZX 


Y=^ 


(12) 


Z^ 


M, 


If  we  make  Vz  =  0,  we  obtain  the  equations  of  the  preceding 


CHAP.  VI.]  ACCELBEATION — EXAMPLES.  67 

article.  If  we  make  x,y,  z  zero,  we  have  the  equations  of  page  37 
if  we  put  V  in  place  of  d. 

General  Analytic  Determination  of  Resultant  Acceleration  and 
Moment  for  Concurring  Component  Accelerations. — The  equations  of 
the  last  two  Articles  hold  good  for  a  point  having  component  accel- 
erations as  well  as  for  component  velocities.  We  have  only  to  sub- 
stitute /  in  place  of  v. 

General  Analytic  Determination  of  Resultant  Displacement  and 
Moment  for  Concurring  Component  Displacements. — The  same  equa- 
tions hold  for  a  point  having  component  displacements.  We  have 
only  to  substitute  d  in  place  of  v.  If  we  then  make  x,  y,  z  =  0  we 
have  the  equations  of  page  37- 


EXAMPLES. 


(1)  A  paint  P  given  by  the  coordinates  x=  +  Bft.,  y  —  +  4:ft., 
z  =  0  has  the  component  velocities  v,  =  40,  Vi  =  50,  Va  =  60  ft.  per 
sec.,  making  the  angles  with  X,  Y  and  Z,  cxi  =  60°,  /i,  =  150%  ri  = 
90°;  a,  =  120%  fJ,  =  30°,  ^^  =90^  a,  =  120  \  /J3  =  150°,  r^  =  90°. 
Find  the  resultant  velocity  and  the  resultant  moment  about  the 
origin. 

Ans.  The  component  velocities  are  in  one  plane  and 

«a;  =  +  20  -  25  -  30  =  -  35  ft.  per  sec; 

%  =  -  34.64  +  43.3  -  51.96  =  -  43.3  ft.  per  sec. 


"The  resultant  ®r  =  V'»x^  +  f>y^  —  55.67  ft.  per  sec,  making  with  the  hori- 
zontal the  angle  cos  a  —  ~  =  pir-^x  or  a  =  128°  57'  17",  and  with  the  vertical 
^  th-       55.67 

angle  cos  6  =  -^  =  '^i^f,  or  6  =  141°  2'  43". 
^  Vr        55.67 

The  moment  of  the  resultant  velocity  Vr  with  reference  to  0  is  Mz  —  —  130 
-f  140  =  4-10  sq.  ft.  per  sec. 

The  direction  of  motion  of  the  radius  vector  in  the  plane  XT  is  therefore 
counter-clockwise. 

The  lever-arm p  =  ^^-^  —  about  0. 18  ft. 

The  equation  of  the  line  of  direction  of  the  resultant  velocity  isy  =  1.237aj 
-(-0.286  ft. 

The  intercepts  on  the  axes  are  y'  —  +0.286  ft.,  of  =  —  0.231  ft. 

(2)  A  point  P  given  by  co-ordinates  x  =  +  3  ft.,  y  =  +  4: ft.,  z  = 
+  5  ft.  has  the  component  velocities  v,  =  40,  Vi  =  50,  Vs  =  60  ft.  per 
sec,  making  the  angles  with  X,  Y,  Z,  »:  =  60°,  /^i  =  100°,  yx  obtuse  ; 
<u-2  =  100°,  fii  =  60°,  r-i  acute  ;  n-3  =  120°,  03  =  100°,  r^  acute.  Find 
the  resultant  velocity  and  the  resultant  moment  about  the  origin. 

Ans.  We  find  the  angles  y  (page  12)  by  the  equation 
cos'^  y  =z  —  cos  (or  -|-  (i)  cos  (a  —  ft). 

Hence  y,  =  148°  2'  31".7,      y^  =  31°  57'  28".3,      y^  =  31°  57'  28".3. 
Bj;  =  -f  20  -  8.6824  -  30  =  -  18.6824  ft.  per  sec. 
ty  —-  6.946  +  25  -  10.419  =  -t-  7.635  ft.  per  sec 
Vz  =-  33.937  +  42.421  +  50.907  .=  +  59.391  ft.  per  sec 


68  KINEMATICS— GENERAL   PBIN0IPLE8.  [CHAP.   VI. 

The  resultant  velocity  is 


«r  =  i/'Vx'  +  «j/*  +  Wz"  =  63.73  ft.  per  sec, 
making  with  the  axes  of  X,  Y,  Z  angles  given  by 
- 18.6834 


63.73 


or  a  =  107°  19'  36"; 


COS  6  =  ^^^2^,       or  &  =    83°     0'  33"  ; 

cosc  =  t|:||^,      or    c=   18"  46'  42". 

The  moments  in  the  co-ordinate  planes  are 

Jf^  =:  +  32.905  +  74.7396  =  +    97.6346  sq.  ft.  per  sec. 
Mx=  -]-  337.564  -  38.175  =  +  199.389     "    "     "     " 
My=  -  93.412  -  178.173  =  -  371.585     "  "     "     " 

The  resultant  moment  is 


Mr  =  V^'"'  +  J^y'  +  Mz    =  350.78  sq.  ft.  per  sec. 
The  line  representative  makes  with  the  axes  of  X,  Y,  Z  angles  given  by 
Jfx       4- 199.389 

COS.  =^  =  ^^11:^^.   ore  =  140-44'    8"; 
Mr  350.78 

eos/=^  =  +  -'-^:^.   or/=    73»  50' 31". 
•'       Mr  350.78  '       '' 

Looking  along  this  line  towards  0,  the  motion  of  the  radius  vector  is 
counter-clockwise. 

The  equations  of  the  projections  of  the  direction  of  the  resultant  velocity  Vr 
upon  the  co-ordinate  planes  are  : 

on  plane  XY,    y  -  -  0.408«  +    5.326  ft.; 

on  plane   FZ.     s  =  +  7.778jr  -  36  115  ft.; 

on  plane  ZX,     x=  -  0.314z  4-    4.573  ft. 

The  point  in  which  the  direction  of  the  resultant  velocity  pierces  the  plane 
Xris  given  by  «'  =  -f  4.573  ft.,  y'  - -\-  3.357  ft. 

The  point  in  which  the  direction  of  the  resultant  velocity  pierces  the  plane 
YZ is  given  by  y'  -=  +  5.336  ft.,  s'  =  -|-  14.56  ft. 

The  point  in  which  the  direction  of  the  resultant  velocity  pierces  the  plane 
ZX  is  given  by  z'  =  -  36.115  ft.,  a;'  =  +  13.788  ft. 

(.3)  A  point  given  hy  the  co-ordinates  x=:  +  3  ft.,  y  =  +  4  ft., 
z  =  0,  has  the  component  accelerations  fi  =  40, /i  =  50,  fa  =  60ft.-per- 
sec.  per  sec.,  making  the  angles  with  X,  Y  and  Z,  ^i  =  60°,  /?i  obtuse, 
Xi  —  90° ;  /32  =  30°,  0-2  obtuse,  y-i  =  90° ;  a^  =  120",  A  obtuse,  Xi  =  90°. 
jPmd  the  resultant  acceleration  and  the  resultant  moment  about  the 
origin. 

Ans.  (page  64).   The  component  accelerations  are  in  one  plane  and 
/a;=  ~f  20  —  25  —  30  =  —  35  ft.-per-sec  per  sec. 
fy=  -  34.64  +  43.3  -  51.96  =  -  43.3  ft.-per  sec.  per  sec. 

The  resultant  fr=  Vf^*  +/""  =  55.67  ft.-per-sec.  per  sec,  making  with 


CHAP.   VI.]  ACCELERATION — EXAMPLES.  69 

the  liorizontal  the  angle  cos  a  —  -^  —  ^,  ^^,  or  a  ~  128°  57'  17",  and  with 

/       50.67 

the  vertical  the  angle  cos  6  =  ^  =    ^,  ^"  ,  or  i*  =  141°  2'  43'. 

/         00.67 

The  moment  of  the  resultant  acceieraiion  with  reference  to  0  is 
Mz  =  —  130  -j-  140  =  +  10  square  feet-per-sec.  per  sec.  The  direction  of  mo- 
tion of  the  radius  vector  in  the  plane  XT  is  therefore  counter-clockwise. 

The  lever-arm  n  of  the  resultant  is  p  =  ;rir-;r;z^  =  about  0.18  ft. 
^  ^      o5.67 

The  equation   of  the  line  of   direction  of   the    resultant   acceleration   is 

y  =  1.287a;  -+-  0.286  ft. 

The  intercepts  on  the  axes  are  y'  =  +  0.286  ft.,  x'  =  —  0.231  ft. 

(4)  A  point  given  by  the  co-ordinates  x  =  +3ft.,  y=+4ft.y 
=  +  5  ft.  has  the  component  accelerations  fi  =  40,  /,  =  50,  /a  =  60 
p;.-per-sec.  per  sec.,  making  the  angles  with  X,  F,  Z,  a\  =  60°, 
fii  =  100°,  Xi  obtuse;  a^  =  100°,  fi^  =  60°,  y^  acute;  az  =  120°, 
fis  =  100°,  xs  acute.  Find  the  resultant  acceleration  and  the  result- 
ant moment  about  the  origin. 

Ans.  (page  65).   We  find  the  angles  y  (page  12)  by 

cos*^  =  —  cos  (or  -)-  (i)  cos  {a  —  /3). 

Hence  y^  =  148°  2'  31  ".7,     y^  =  31°  57'  28".3,     r»  =  31°  57'  28".3. 

/a,  =  +  20  -  8.6824  -  30  =  -  18  6824  ft.-per-sec.  per  sec; 

/„=- 6.946  +  25- 10.419  =+  7.635     "    "     "      "     "    ; 

/2  =  - 83. 937 +  42. 421 +  50. 907  =  +  59. 391     "    "     "      "     «   . 

The  resultant  acceleration  is 


fr  =  Vfx^  +fv  +f''  =  62.73  ft.-per-sec.  per  sec., 
making  with  the  axes  of  X,  Zand  Wangles  given  by 

-  18.6824  -.nr^o  .„,  ofl,. 

cos  a  =  —     ,  or  a  =  107  19  36' ; 
cos6=i|;^.  or  6=  83°  0' 33"; 

cos  c  =  i^^.  or  e  =   18°  46'  42". 

The  moments  in  the  co-ordinate  planes  are 

Mz  =  -{-    22.905+74.7296  =  +    97.6346  sq.  ft.-per-sec.  per  sec. 
Jfx  =  + 237.564- 38.175    =  +  199.389     "     "     "    "      "    " 
My=-    93.412  - 178.173  =- 271.585    "     "     "     "      '     " 

The  resultant  moment  is 


Mr  =  V ^x" -\- Mv"* -{- Mz^  =  350.78  sq.  ft.-per-sec.  per  sec. 

The  line  representative  makes  with  the  axes  of  X,  7,  Z  angles  ^ven  by 

,      Mx       +199.889  -        K^oni,or,n 

cosd  =  ^=  ^Q^e     .  or  d=   55°  21' -37"; 

My       -271.585  ..noAA>    a" 

«°^^  =  :#  =  -407:6-''''^  =  ^^^  ^    ^  ' 

Mz  _  +97.6346 
cos/==  -^  -      407  g    .  or  /_    73   50  21  . 


70  KINEMATICS — GENERAL   PRINCIPLES.  [CHAP.    VI. 

Looking  along  this  line  towards  0,  the  motion  of  the  radius  vector  is  coun- 
ter-clockwise. 

The  equations  of  the  projection  of  the  direction  of  the  resultant  acceleration 
/  upon  the  co-ordinate  planes  are  : 

on  plane  XT,  y  =  -  0.408a!  -f  5.226  ft.; 
on  plane  YZ.  z  =  +  7  778^^  -  26.115  ft.; 
on  plane  ZX,     x=-  0.314z  -f  4.572  ft. 

The  point  in  which  the  direction  of  the  resultant  acceleration  pierces  the 
plane  XFis  given  hy  x  = -]-  4.572  ft.,  y'  =  +3.357  ft. 

The  point  in  which  the  direction  of  the  resultant  acceleration  pierces  the 
plane  YZis  given  hyj/=-\-  5.226  ft.,  g'  =  -  14.56  ft. 

The  point  in  which  the  direction  of  the  resultant  acceleration  pierces  the 
plane  ZX  is  given  by  2'  =  —  26.115  ft.,  a;'  =  +  12.788  ft.    . 

(5)  A  point  given  by  the  co-ordinates  x  =  +  3  ft.,  y=  +  4  ft., 
2  =  0  has  the  component  displacements  d\  =  40  ft.,  d^  =  50  ft., 
d3  =  QOft.,  m,aJcing  the  angles  with  X,  Y  and  Z,  a,  ==  60°,  /3,  obtuse, 
yi  =  90'';  /ia  =  30°,  a^  obtuse,  r^  =  90;  as  =  120°,  /ia  obtuse,  y^  =  90°. 
Find  the  resultant  displacement  and  moment  about  the  origin. 

(6)  A  point  given  by  the  co-ordinates  x  =  +  B  ft.,  y  =  +  4  ft., 
z  —  +5ft.  has  the  component  displacements  di  =  4S)ft.,  d^  =  50/^., 
da  —  GO  ft.,  making  the  angles  with  X,  Y  and  Z,  ai  =  60°,  fii  =  100°, 
yi  obtuse;  ^a  =  100°,  /Ja  =  60°,  r-i  acute;  «3  =  120°,  ^3  =  100°,  rs  acute. 
Find  the  residtant  displacement  and  moment  about  the  origin. 


UA-<ir//y—  -^-O'^^ 


< 

•c 


AT  '     /\^CO  . 


CHAPTEE  VII. 

ANGULAR  REVOLUTION   OF  A  POINT.    ANGULAR  SPEED. 


KATE  OF  CHANGE  OF  ANGULAK  SPEED.  EQUATIONS  OF  MOTION  OF  A  POINT 
UNDER  DIFFERENT  RATES  OF  CHANGE  OP  ANGULAR  SPEED.  ANGULAR 
SPEED  IN  TERMS  OF  LINEAR  VELOCITY,  RATE  OF  CHANGE  OF  ANGULAR 
SPEED  IN  TERMS  OF  LINEAR.  MOMENT  OF  LINEAR  VELOCITY  IN  TERMS 
OF  ANGULAR  SPEED.  MOMENT  OP  TANGENTIAL  ACCELERATION  IN  TERMS 
OF  RATE  OF  CHANGE  OP  ANGULAR  SPEED.  NORMAL  ACCELERATION  IN 
TERMS  OF  ANGULAR  SPEED.      MOTION  IN  A  CIRCLK. 

Angular  Revolution  of  a  Point  about  a  Given  Point.— When  a 
point  moves  in  any  path  whatever  from  the  initial  position  Pi  to 
the  final  position  P^  in  any  given  time,  we  have  called  the  distance 
PiPa  the  linear  displacement  (page  34). 

If  we  choose  any  point  in  space  O  as  a  pole 
and  draw  the  radius  vector  OPi  to  the  initial 
and  OP-i  to  the  final  position,  we  call  the  angle 
P,  OP-i  =  0  the  angular  revolution  of  the  point 
P about  O. 

Since  the  angle  9  is  measured  in  radians,  it 
is  independent  of  the  length  of  the  radius  vec- 
tor, or  the  distance  of  P,  and  Pi  from  O  (page  5). 

It  is  also  independent  of  the  position  of  the  plane  of  revolution 
Pi  OPi  in  space,  or  of  the  direction  in  space  of  the  angular  revolu- 
tion. 

It  has,  however,  magnitude  and  sign  (+)  or  (— ),  according  as  the 
radius  vector  moves  in  this  plane  in  one  direction  or  the  other. 

Angular  revolution  has  then  magnitude  and  sign,  but  not  direc- 
tion. It  is  therefore  a  scalar  quantity  like  distance  described  by  a 
point,  and  cannot  be  represented  by  a  straight  line. 

The  student  must  not  confound  angular  revolution  with  "angular  displace- 
ment," which,  as  we  shall  see  hereafter  (page  170),  has  like  linear  displacement, 
direction  as  well  as  sign  and  magnitude,  and  is  therefore  a  vector  quantity 
which  can  be  represented  by  a  straight  line. 

Angular  Revolution  of  a  Point  about  a  Given  Axis.— The  angular 
revolution  in  any  given  time  of  a  moving  point  about  a  given  line 
or  axis  is  the  angle  between  perpendiculars  from  the  initial  and 
final  positions  of  the  point  to  the  axis. 

Thus  let  OA  be  a  given  axis,  Pi  and  Pn 
the  initial  and  final  positions  of  the  mov- 
ing point,  and  PxB,  PaC  perpendiculars  to 
OA.  Then  the  angle  between  PiB  and 
P2C  is  the  angular  revolution  about  OA, 
whatever  the  path  between  P,  and  P^. 

This  angle  is  the  same  as  the  angle 
pCPi,  if  we  complete  the  rectangle  CPi. 
As  the  straight  line  pPi  is  thus  the  pro- 

71 


72  KINEMATICS — GENERAL   PRINCIPLES.  [CHAP.   VII, 

jection  of  the  line  PiPi  on  the  plane  PiCp,  we  see  that  the  angular 
revolution  about  the  axis  is  the  angular  revolution  of  the  projection 
p  of  Pi  about  the  point  C. 

Mean  Angular  Speed  of  a  Point  about  a  Given  Point  or  Axis.— The 
angular  revolution  per  unit  of  time  is  the  mean  angular  speed  of  a- 
point  about  a  given  point  or  axis. 

Like  angular  revolution  it  has  then  magnitude  and  sign  accord- 
ing to  direction  of  motion  in  the  plane  of  revolution,  but  is  inde- 
pendent of  the  position  in  space  of  that  plane.  It  is  therefore  a 
scalar  quantity  like  linear  speed  (page  16). 

When  the  mean  angular  speed  varies  with  the  time  it  is  variable. 
When  it  has  the  same  magnitude  no  matter  what  the  interval  of 
time  it  is  uniform.  A  point  moving  with  uniform  angular  speed  evi- 
dently describes  equal  angles  in  equal  times. 

Instantaneous  Angular  Speed  of  a  Point  about  a  Given  Point 
or  Axis.  —The  limiting  value  of  the  mean  angular  speed  when  the 
interval  of  time  is  indefinitely  small  is  the  instantaneous  angular 
speed.  If  the  instantaneous  angular  speed  is  variable,  the  mean 
angular  speed  has  different  values  for  equal  intervals  of  time. 

The  term  angular  speed  always  signifies  instantaneous  angular 
speed  unless  otherwise  specified. 

Angular  speed  like  mean  angular  speed  is  therefore  a  scalar 
quantity,  having  magnitude  and  sign  according  to  the  direction  of 
motion  in  the  plane  of  revolution,  but  independent  of  the  position 
of  this  plane  in  space. 

The  student  must  not  confound  angular  speed  with  "angular  velocity," 
which,  as  we  shall  see  hereafter  (page  174),  has  direction  as  well  as  sign  and 
magnitude  and  is  therefore  a  vector  quantity. 

Numeric  Equations  of  Angular  Speed.— The  unit  of  angular  speed 
is  evidently  one  radian  per  second.  We  denote  the  magnitude  of 
the  angular  speed  thus  measured  by  the  letter  go. 

If  then  &,  is  the  angle  measured  in  the  plane  of  revolution  from 
any  fixed  line  to  the  initial  position  of  the  radius  vector  and  0  to  the 
final  position  of  the  radius  vector,  we  have  for  the  mean  angular 
speed 


t 


When  the  interval  of  time  is  indefinitely  small,  we  have  in/the 
Calculus  notation,  for  the  instantaneous  angular  speed, 

dO 

•'=di (2> 

Sign  of  Angular  Speed. — These  equations  are  precisely  the  same 
as  equations  (1)  and  (2),  page  16,  simply  substituting  6  for  s.  The 
sign  follows  the  same  rule.  Thus  when  the  angle  is  increasing  the 
value  of  o)  is  positive  (4-),  and  when  decreasing  it  is  negative  (— ); 

Equation  (1)  is  thus^  general  if  we  take  angles  in  any  one  direc- 
tion in  the  plane  of  revolution  measured  from  a  fixed  line  in  that 
plane  as  positive,  and  in  the  opposite  direction  as  negative. 

Angular  speed,  then,  whether  uniform  or  variable,  mean  or  in- 
stantaneous, is  independent  of  direction  in  space.  It  is  entirely 
comparable  to  linear  speed  (page  15). 

Change  of  Angular  Speed. — When  the  angular  speed  of  a  pomt 
varies,  the  difference  between  the  final  and  initial  instantaneous 
speeds  for  any  interval  of  time  is  the  integral  change  of  angular 
speedy 


CHAP.  VII.]       RATE   OF   CHANGE    OF   ANGULAR   SPEED.  75 

Mean  Rate  of  Change  of  Angular  Speed. — The  integral  change  of 
angular  speed  per  unit  of  time  is  the  mean  rate  of  change  of  angu- 
lar speed. 

When  the  mean  rate  of  change  varies  with  the  time  it  is  vari- 
able. When  it  has  the  same  magnitude  no  matter  what  the  inter- 
val of  time  it  is  uniform. 

Instantaneous  Rate  of  Change  of  Angular  Speed.— The  limiting 
value  of  the  mean  rate  of  change  of  angular  speed  when  the  inter- 
val of  time  is  indefinitely  small  is  the  instantaneous  rate  of  change 
of  angular  speed. 

Rate  of  change  of  angular  speed  should  always  be  understood  as 
meaning  instantaneous  rate  of  change  unless  otherwise  specified. 

Rate  of  change  of  angular  speed  may  be  zero,  uniform  or  vari- 
able. When  it  is  zero  the  angular  speed  is  uniform  and  the  same 
as  the  mean  speed  for  any  interval  of  time. 

When  it  is  uniform  the  rate  of  change  of  angular  speed  is  the 
same  as  the  mean  rate  of  change  for  any  interval  of  time. 

When  it  is  variable  the  mean  rate  of  change  has  different  values 
for  equal  intervals  of  time. 

Numeric  Equations  of  Rate  of  Change  of  Angular  Speed. — The 
unit  of  rate  of  change  of  angular  speed  is  then  one  radian-per-sec. 
per  sec.     We  denote  its  magnitude  thus  measured  by  the  letter  a. 

If  then  00 1  is  the  initial  and  co  the  final  instantaneous  angular 
speed,  we  have  for  the  mean  rate  of  change  of  angular  speed 

GO  —  Gi3i 

"  =  -7-- <»> 

anmf  or  the  instantaneous  rate  of  change  of  angular  speed 

"=d^  =  d^ ^^> 

Cf 

Sign  of  Rate  of  Change  of  Angular  Speed. — These  equations  are 
precisely  the  same  as  equations  (1)  and  (2),  page  25,  simply  substi- 
tuting 00  for  V  and  6  for  s  and  a  for  a.  The  value  then  of  a  is  posi- 
tive (  + )  when  the  angular  speed  increases  and  negative  (— )  when 
it  decreases  during  the  time. 
..i-.  •  It  is  tevident  that  this  sign  has  no  reference  to  position  in  space. 
Rate  of  change  of  angular  speed  is  then  a  scalar  quantity. 

The  student  must  not  confound   rate  of  change  of  angular  speed  with 
,y  angular  acceleration,"  which,  as  we  shall  see  hereafter  (page  175),  has  direc- 
tion as  well  as  sign  and  magnitude  and  is  therefore  a  vector  quantity. 

Equations  of  Motion  of  a  Point  under  Different  Rates  of  Change 
of  Angular  Speed. — We  have  equations  preciseljr  similar  to  those 
for  linear  speed,  page  27.  We  have  only  to  substitute  a?  for  v,  a  for 
a,  5  for  s. 

(a)  Rate  of  Change  of  Angular  Speed  Zero. — In  this  case  if  Qi  is 
the  initial  angle  measured  in  the  plane  of  revolution  of  the  radius 
vector  from  a  fixed  line  in  that  plane,  and  6  the  final  angle,  we  have 

6  —  Oi                    .        ^        «  /-4  ^ 

GO  = ,     or     cot  =  6  —  Qi (1) 

Revolution  in  any  one  direction  in  the  plane  being  taken  as  posi- 
tive, in  the  other  direction  it  is  negative.  Then  if  &?  comes  out  (-f-) 
it  denotes  revolution  in  the  assumed  positive  direction;  if  (— ),  in 
the  opposite  direction.  If  t  comes  out  (+)  it  denotes  time  after,  if 
(— )  time  before,  the  start. 


74  KINEMATICS — GENERAL   PRINCIPLES.         [CHAP.    VII. 

(b)  Rate  of  Change  of  Angular  Speed  Uniform. — When  the  rate 
of  change  of  angular  speed  is  uniform,  the  instantaneous  rate  of 
change  of  angular  speed  at  any  instant  is  equal  to  the  mean  rate  of 
■change  for  any  interval  of  time. 

If  fi9i  and  GO  are  the  initial  and  final  instantaneous  angular  speeds, 
we  have  then  for  the  rate  of  change  of  angular  speed 

GO  —   COi 

"  =  -!-- «> 

The  value  of  a- is  (+)  when  the  angular  speed  increases  and  (— ) 
when  it  decreases  during  the  time. 

From  equation  (2)  we  have 

GO  =  GOi  +  at (3) 

The  average  angular  speed  is 

2 —  =  °^'  +  2 ^^ 

The  angle  described  in  the  time  t  is 

Q  -  0,  =  .^L±^t  =  cod  +  lat\       ...     (5) 

Inserting  the  value  of  t  from  (2)  'we  have 

»-»'=-^- (« 

Hence  oj"  =  gox^  +  2a{^  —  0,) (7) 

In  applying  these  formulas,  a  is  positive  (+)  when  the  angular 
speed  increases  and  negative  (— )  when  it  decreases  during  the 
time,  without  regard  to  direction  of  revolution. 

If  angles  6,  Oi  in  one  direction  are  taken  as  (+),  angles  in  the 
opposite  direction  are  (— ). 

Angular  speeds  aj,  (Wi  are  positive  (+)when  motion  is  in  the 
assumed  positive  direction,  and  negative  (— )  when  in  the  other 
direction.  A  positive  t  denotes  time  after  the  beginning  of  motion, 
and  a  negative  t  time  before. 

[(c)  Rate  of  Change  of  Angular  Speed  Variable] — If  the  rate  of  change  of 
angular  speed  is  variable,  we  have  from  (1),  in  Calculus  notation, 

'^  =  av (8) 

doo       d^B 

^=dt=d^' (9) 

and  from  (8), 

0  -  Oi  =  /  oodt (10) 

The  preceding  equations  can  be  directly  deduced  from  these  as  on  page  28^ 
Angular  Speed  in  Terms  of  Linear  Velocity. — If  a  point  moves 
A  from  Pi  to  Pi  in  any  path  in  the  time  t,  the 

^  linear  displacement  in  that  time  is    the 

"    ^  .Rj     chord  P1P2. 

Pj^i- '(py^  If  we  take  O  as  a  pole,  the  angular  dis- 

placement of  revolution  is  P1OP2  =  9.   The 
mean  linear  velocity  is  :^ 

'^  chord  P^P^  ,  .  / 

and  the  mean  angular  speed  is  —.  n 


CHAP.  VII.]  ANGULAR  SPEED  IN  TERMS  (HP  LINEAR  VELOCITY.  75 

From  Pi  draw  PiN  perpendicular  to  OP^i.  Then  if  angle  PiP^N 
=  0,  we  have 

PiN=  chord  PiPi .  sin  0. 

Dividing  by  t,  we  have  / 

PiN  _  chord  PiPa .  sin  (f> 

t     ~  t  '"' 

But  PiN  =  r  sin  0,  where  r  is  the  radius  vector  OPi.    Hence 

rsin6  _  chord  Pi  Pa .  sin  0 

t      ~'  i  ■ 

If  now  the  time  is  indefinitely  small, -—- — ?  becomes  the  in- 

I 
stantaneous  velocity  v,  and  cp  becomes  the  angle  APiV  =  e  between  ; 
the  radius  vector  OPi  and  the  instantaneous  velocity  v,  and  sin  9  ; 

becomes  S,  and  —becomes  the  instantaneous  angular  speed  oo.  Hence  j 

usine  „. 

rcj  =  V  sm  e,    or    go  = . (1) 

r 

In  general,  then,  whatever  the  path  or  wherever  the  pole. 
The  magnitude  of  the  angular  speed  at  any  point  is  equal  to  the 
magnitude  of  the  component  of  the  linear  velocity  at  that  point  per- 
pendicular to  the  radius  vector,  divided  by  the  magnitude  of  the 
radius  vector. 

If  the  pole  O  is  taken  at  the  centre  of  curvature,  so  that  OPi  is 
equal  to  the  radius  of  curvature  p,  then  e  =  90°  and  we  have  poo  =  v 

v 
or  o)  =  — . 
P 

Rate  of  Change  of  Angular  Speed  in  Terms  of  Tangential  Linear 
Acceleration. — If  ft  =  a  is  the  magnitude  of  the  linear  tangential 
acceleration  or  rate  of  change  of  speed  at  any  point,  then  we  can 
prove,  precisely  as  in  the  preceding  Article,  that 

a  =  fi^B:I (2) 

r 

where  a  is  the  magnitude  of  the  rate  of  change  of  angular  speed. 

Hence  the  magnitude  of  the  rate  of  change  of  angular  speed  at 
any  point  is  equal  to  the  magnitude  of  the  component  of  tJie  linear 
tangential  acceleration  at  that  point  perpendicular  to  the  radius 
vector,  divided  hy  the  magnitude  of  the  radius  vector. 

If  the  pole  O  is  taken  at  the  centre  of  curvature,  e  =  90°  and  we 

have  pa  —  ft  or  a  =  — . 

Moment  of  Linear  Velocity  in  Terms  of  Angular  Speed.— We  can 
resolve  the  linear  velocity  v  at  the  point  P  into  a  component  v  cos  e 
along  the  radius  vector  and  a  component  ^ 

v  sin  e  perpendicular  to  the  radius  vector.  Xg 

The  moment  of  the  first  relative  to  the  pole       ^ -^Z^ — Uv 

is  zero.     Since  the  moment  of  r  is  equal  to       ;         /      \"~"~"^ 
the  algebraic  sum  of  the  moments  of  its  com-     ^j        Z        \     ^ 
ponents  (page  62),  if  we  take  moments  about       !    / 
the  pole,  we  have  ]/^ 

o 
t}p  =  V  sm  € .  r. 


76  KINEMATICS — GENERAL    PRINCIPLES.  [CHAP.    VII. 

But  we  have  just  seen  that  v  sin  e  =  roo.    Hence    . 

vp  =  r'^oo (3) 

That  is,  the  magnitude  of  the  moment  of  the  linear  velocity  at  any      /-, 
point  relative  to  the  pole  is  equal  to  the  magnitude  of  the  angular    /  f  ]/t/* 
speed  at  that  point,  multiplied  by  the  square  of  the  magnitude  of  the   k 
radius  vector. 

Since  v  sin  e  is  the  normal  component  of  v,  v  sin  e  .  r  is  twice  the 
areal  velocity  of  the  radius  vector  (page  61). 

Moment  of  Linear  Tangential  Acceleration  in  Terms  of  Rate  of 
Change  of  Angular  Speed. — We  can  resolve  the  tangential  accelera- 
tion /( into  components  along  and  perpendicular  to  the  radius  vector 
and  thus  obtain,  precisely  as  in  the  preceding  Article, 

fP=ftPt='^''' (4) 

Hence  the  magnitude  of  the  moment  of  the  linear  tangential  ac- 
celeration at^  any  point  relative  to  the  pole  is  equal  to  the  magnitude 
of  the  rate  of  change  of  angular  speed  at  that  point,  multiplied  by 
the  square  of  the  magnitude  of  the  radius  vector. 

Since  ft  sin  e  is  the  normal  component  of  ft ,  ft  sin  e  .r  =  r''a  is 
twice  the  areal  acceleration  of  the  radius  vector  (page  61). 

Normal  Linear  Acceleration  in  Terms  of  Angular  Speed. — We 
have  seen  (page  53)  that  when  a  point  moves  in  any  path,  the 

v'^ 
magnitude  of  the  normal  acceleration  f^  is  given  by  f^  =  — ,  where 

p  is  the  radius  of  curvature. 

If  we  take  the  pole  at  the  centre  of  curvature,  then,  we  have  v  — 
pa,  and  hence 

/n  =  77  =  P*"'  =  ^'^ (5) 

The  magnitude  of  the  normal  linear  acceleration  at  any  point  is 
equal  to  the  magnitude  of  the  radius  of  curvature  at  that  point, 
multiplied  by  the  magnitude  of  the  square  of  the  angular  speed,  or  to 
the  velocity  voa  in  the  hodograph  (page  52). 

Since  for  any  path  roo  =  v  sin  e,  we  have 


Hence,  in  general, 


roo 

sin  e 


.  .      V'  r'ftj' 

f     =VGO  =  pCO^  =  —  = 


P        p  sm"  € 

where  r  is  any  radius  vector  when  the  pole  is  not  at  the  centre  of 
curvature,  and  e  is  the  angle  of  v  with  this  radius  vector. 

Motion  in  a  Circle. — For  a  point  moving  in  a  circle  we  have 
e  =  90°  and  r  =  p.    Hence  from  (1),  page  75,  we  have 

V 

v  =  roo,     or    00=  — (1> 

r 

If  r  is  unity,  we  have  the  numeric  equation  go  =  v,  that  is,  the 
number  of  units  of  angular  speed  is  equal  to  the  number  of  units  of 
linear  speed  at  distance  unity. 

From  (2),  page  75,  we  have 

ff  =  ra,     or    a=  — (2) 


-CHAP.    VII.]  EXAMPLES — ANGULAR   SPEED. 

If  r  is  unity,  we  have  the  numeric  equation  a  =ft,  that  is,  the 
number  of  units  of  rate  of  change  of  angular  speed  is  equal  to  the 
number  of  units  of  linear  tangential  acceleration  at  distance  unity. 

From  (5),  page  76,  we  have  in  any  case 

fn  =  'f^^^ (3) 

or  the  normal  linear  acceleration  is  equxil   to  the  velocity  in  the 
hodograph  (page  53).     Inserting  the  value  of  v  from  (1), 

fn  =  roo'=~^    or    c.^=-^ (4) 

If  r  is  unity,  we  have  the  numeric  equation  fn  =  v'^  =  oo^,  that  is, 
the  number  of  units  of  the  normal  linear  acceleration  is  equal  to  the 
square  of  the  number  of  units  of  linear  velocity  at  distance  unity  ; 
or  the  square  of  the  number  of  units  of  angular  speed  is  equal  to  the 
number  of  units  of  the  normal  linear  acceleration  at  distance  unity. 

We  have  also  for  the  total  resultant  linear  acceleration 

/  =   Vfi'  +fn\ (5) 

Since  the  component  fn  passes  through  the  centre,  the  moment 
of  /  relative  to  the  centre  is  equal  to  the  moment  of  /^.    Hence 

w  —  r^co, (6) 

fp=ftr  =  r'a (7) 

give  the  moments  of  v  and  /  with  respect  to  the  centre. 

If  the  point  starts  from  rest  and  acquires  the  velocity  v  in  the 

V 

time  t,  under  constant  tangential  acceleration,  we  have  /^  =  — , 

GO 

a  =  —. 

t 

Graphic  Representation  of  Rate  of  Change  of  Angular  Speed. — 
We  can  represent  intervals  of  time  by  distances  laid  oflE  horizontally 
and  the  corresponding  angular  speeds  by  distances  laid  off  vertically 
and  thus  obtain  the  same  diagrams  as  for  linear  speed  given  on 
page  29. 

EXAMPLES. 

(1)  The  angular  speed  of  a  point  moving  in  a  plane  about  some 
assumed  point  changes  from  50  to  30  radians  per  sec.  in  passing 
through  80  radians.  Find  the  constant  rate  of  change  of  angular 
speed  and  the  time  of  motion. 

ca'  —  aji'  ,.  mi  .  .         T 

Ans.  a  =  -—- ^— -  =  —  10  radians-per-sec.  per  sec.     The  minus  sign  de- 

2(e  —  6i)  f  r 

notes  decreasing  speed,     t  = =  3  sec. 

(2)  Draic  a  figure  representing  the  motion  in  the  preceding 
example,  and  deduxie  the  results  directly  from  it. 

Ans.  Average  speed  =  — '^ —  =  40  radians 

per  sec.     Hence  40i  =  80  or  <  =  3  seconds.    Also       '•'r*  ^ 

30-50  ,_      ,.  '^-*^ 

a.  = =  —  10  radians-per-sec.  per  sec. 


78  KINEMATICS — GENERAL   PRINCIPLES.  [CHAP.    VII» 

(3)  A  point  mooing  in  a  plane  has  an  initial  speed  of  60  radians 
per  sec.  about  an  assumed  point  and  a  rate  of  change  of  speed  of 

+  40  radians-per-sec.  per  sec.    Find  the  speed  after  8  sec;  the  time 
required  to  describe  300  radians  ;  the  change  of  speed  ivhile  describ- 
ing that  angle;  the  final  speed. 
Ans.  See  Example  (9),  page  '31. 

(4)  If  the  motion  in  the  last  example  is  retarded,  find  (a)  the 
angular  revolution  from  the  start  to  the  turning-point ;  (b)  the  angle 
described  from,  the  start  after  10  sec. ;  the  speed  acquired  and  the 
angle  between  the  final  and  initial  positions  ;  (c)  the  angle  described 
during  the  time  in  which  the  speed  changes  to  —  90  radians  per  sec, 
atid  this  time  ;  id)  the  time  required  by  the  moving  point  to  return 
to  the  initial  position. 

Ans.  See  Example  10,  page  32. 

(5)  A  point  moving  in  a  plane  describes  about  a  fixed  point 
angles  of  120  radians,  228  radians  and  336  radians  in  successive 
tenths  of  a  second.  Show  that  this  is  consistent  with  uniform  rate 
of  change  of  angular  speed,  and  find  this  rate. 

Ans.   a  —  10800  radians-per-sec.  per  sec. 

(6)  Two  points  A  and  B  move  in  the  circumference  of  a  circle 
with  uniform  angular  speeds  go  and  go'.  The  angle  between  them  at 
the  start  is  a.  Find  the  time  of  the  nth  meeting,  the  angles  described 
by  A  and  B,  and  the  interval  of  time  between  two  successive  meet- 
ings. 

Ans.  See  Example  (21),  page  21. 

Time  of  the  wth  meeting,  tn  = ' ; • 

CO  ±    GO 

Angle  described  by  .4  is  6  =  cotn. 

"         "  S  is  T  a. 
Interval  of  time  between  two  successive  conjunctions  is 

\A'  '    _^-/~r  CO  ±  GO 

where  we  take  the  (+)  or  (— )  sign  for  a  according  as  B  is  in  front  of  or  behind 
A  at  start,  and  {-\-)  or  (— )  sign  for  go'  according  as  the  points  move  in  opposite 
or  the  same  directions. 

(7)  What  is  the  angular  speed  of  a  fly-wheel  5  ft.  in  diameter 
ivhich  makes  30  revolutions  per  minute,  and  what  is  the  linear  ve- 
locity of  a  point  on  its  circumference  f  Also  find  its  linear  normal 
acceleration  and  the  moment  of  its  velocity  with  reference  to  the 
centre. 

Ans.  Tt  radians  per  sec;  2.57r  ft.  per  sec,  tangent  tocirc;  2.5;r'^  ft.-ptr- 
sec.  per  sec;  6.25;r  sq.  ft.  per  sec. 

(8)  Find  the  linear  and  angular  speed  of  a  point  on  Hie  earth's 
equutor,  taking  radius  4000  miles ;  also  the  linear  normal  accelera- 
tion. 

Tt 

Ans.  1535.9  ft.  per  sec;  -—  radians,  or  15°  per  hour;  0.112  ft.-per-sec  per 
sec. 

"%.    (9j  The  angular  speed  of  a  wheel  is  —Tt  radians  per  sec.    Find  the 

linear  speed  of  points  at  a  distance  of  2ft.,  4  ft.  and  10  ft.  from  the 
centre,  also  the  linear  normal  acceleration. 


■^k. 


i, 


CHAP.   VII.]  EXAMPLES — ANGULAR  SPEED.  79 

g 
Ans.  ^  ic,  Zic,  7. 5;r  ft.  per  sec. 

a 

9         9        45 

^-nr",   -j-ffS   -^-tt'^  ft.-pei-sec.  per  sec. 
o  4  o 

(10)  ijf  the  linear  speed  of  a  point  at  the  equator  is  v,  find  the 
speed  linear  and  angular  at  any  latitude  A. 

n 
Ans.  «  cos  A  ;  -—  radians  per  liour,  or  15°  per  hour. 

(11)  A  point  moves  with  uniform  velocity  v.  Find  at  any  instant 
its  angular  speed  about  a  fixed  point  whose  distance  from  the  path 
is  a. 

Ans.  -i- radians  per  sec,  where  r  is  the  radius  vector.    Uniform  velocitr 
r" 

means  uniform  speed  in  a  straight  line  (page  43). 

Hence  the  angular  speed  of  a  point  moving  with  uniform  speed  in  a  straight 

line  is  inversely  proportional  to  the  square  of  the  distance  of  the  point  from  a 

fixed  point  not  in  the  line. 

(12)  The  speed  of  the  periphery  of  a  mill-wheel  12  feet  in  diame- 
ter is  6  feet  per  sec.  How  many  revolutions  does  the  wheel  m^xke 
per  sec.  f 

Ans.  jr —  revolutions. 

(13)  The  time  i^  between  5  and  6  d'cloch,  and  the  hour  and  min- 
ute hands  are  together.     What  is  the  time  f 

Ans.  oh.  27  m.  16  sec.  (see  Example  (6) ). 

(14)  Express  in  degrees  and  radians  the  angle  made  by  the  hands 
of  a  clock  af  3.35  o'' clock. 

Ans.  102.5  deg. ,  1.79  radians. 

(15)  Find  the  multiplier  for  changing  revolutions  per  minute  into 
radians  per  second. 

Ans.  0.10472  rad.  per  sec.  =  1  rev.  per  min. 

(16)  The  minute  and  second  hands  point  in  the  same  direction  at 
12  o'clock.     When  do  they  next  point  in  the  same  direction  f 

Ans.  1  min.  Ig'^  sec.  after  twelve.     (See  Example  (6)). 

(17;  Two  clocks  are  together  at  XII.  When  the  first  comes  to  /, 
it  has  lost  a  second ;  when  the  second  comes  to  I,  it  has  gained  a 
second.    How  far  are  they  apart  in  12  hours  f 

Aus.  24  sees. 

ik        (18)  Two  men  start  together  to  walk  around  a  circular  course,  one 
taking  75  minutes  to  the  round,  the  other  90.     When  will  they  be  to- 
gether again  at  the  starting-point  f 
Ans.  7.5  hours.     (See  Example  (6)). 

\        (19)  Tlie  hour-hand  of  a  watch  is  f  of  an  inch  long,  the  minute- 
'^'  hand  f  of  an  inch,  and  the  second-hand  i  of  an  inch.     Compare  the 
lineal  speeds  of  their  points  and  the  angular  speeds. 

Ans.  5  :  112  :  2800;     1  :  12  :  720. 

(20)  Deduce  the  equivalent  of  longitude  for  one  minute  of  time 
and  for  one  second  of  time. 
Ans.  15'  to  1  min.,   15"  to  1  sec. 


so  KINEMATICS— GEKERAL  PRINCIPLES.  [CHAP.  VII. 

(21)  TTie  diameter  of  the  earth  is  nearly  8000  miles.     Required 
the  circumference  at  the  equator  and  the  linear  speed  at  latitude  60°. 
Ans.  25000  miles;  521  miles  per  hour. 

' '  (22)  The  wheel  of  a  bicycle  is  52  inches  in  diameter  and  performs 

5040  revolutions  in  a  journey  of  65  minutes.  Find  the  speed  in 
miles  per  hour  ;  the  angular  speed  of  any  point  about  the  axle  ;  the 
areal  velocity  of  a  spoke ;  the  relative  velocity  of  the  highest  point 
with  respect  to  the  centre. 

Ans.  12  miles  per  hour;  8.12  radians  per  sec;  19.06  sq.  ft.  per  sec;  12 
miles  per  hour. 

(23)  In  going  120  yards  the  front  wheel  of  a  carriage  maJces  six 
revolutions  more  than  the  hind  wheel.  If  each  circumference  were  a 
yard  longer.,  it  would  make  only  4  revolutions  more.  Find  the  cir- 
cumference of  each  wheel. 

Ans.  4  yards  and  5  yards. 

(24)  If  the  velocity  of  a  point  is  resolved  into  several  components 
in  one  plane,  show  that  its  angular  speed  about  any  fixed  point  in 
the  plane  is  the  sum  of  the  angular  speeds  due  to  the  several  compo- 
nents. 

(25)  A  point  moves  with  uniform  speed  v  in  a  circle  of  radius 

r.    Show  that  its  angular  speed  about  any  point  in  the  circumfer- 

.     V 
ence  ts  — . 
2r 

(26)  Show  that  the  angular  speed  of  the  earth  about  the  sun  is 
proportional  to  the  apparent  area  of  the  sun's  disk.  [The  radius 
vector  from  the  sun  to  the  earth  sweeps  over  equal  areas  in  equal 
times.] 

W  (^  (27)  A  point  P  moves  in  a  parabola  with  constant  angular  speed 

\i  ii\     about  the  focus  S.    Show  that  its  linear  speed  is  proportional  to 

tl-i.    ySP- 

^.L  V        (28)  A  point  starting  from  rest  moves  in  a  circle  with  a  constant 

rate  of  change  of  angular  speed  of  2  radians -per-sec.  per  sec.     Find 

'j»^       the  angular  speed  at  the  end  of  20  sec.  and  the  angular  displacement 

^         of  revolution ;    also  the  linear  speed  and  distance  described  and 

Si  -^      the  number  of  revolutions  ;  also  the  linear  tangential  acceleration 

r         and  the  normal  linear  acceleration  at  the  end  of  20  sec. 

Ans.  40  rad.  per  sec;  400  radians;  40r  ft.  per  sec;  400rft. ;  - —  revolu- 
tions; 2r  ft.-per-sec.  per  sec.  tangential  acceleration;  1600r  ft.-per-sec  per  sec. 
normal  acceleration. 

(29)  A  point  moving  with  uniform  rate  of  change  of  angidar 
speed  in  a  circle  is  found  to  revolve  at  the  rate  of  8i  revolutions  in 
the  eighth  second  after  starting  and  7i  revolutions  in  the  thirteenth 
second  after  starting.  Find  its  initial  angular  speed  and  its  uni- 
form rate  of  change  of  angular  speed  ;  also  the  initial  linear  speed 
and  rate  of  change  of  speed;  also  the  initial  normal  acceleration. 

Ans.  20.27r  radians  per  sec;  —  0.47r  radians-per-sec  per  sec;  20.2;rr  ft. 
p^sec;  —  0.47rr  ft.-per-sec.  per  sec;  408.047rV  ft.-per-sec  per  sec. 

ly  (30)  A  point  starts  from  rest  and  moves  in  a  circle  with  a  uni- 
form rate  of  change  of  angular  speed  of  18  radians-per-sec.  per  sec. 
Find  the  time  in  which  it  makes  the  first,  second  and  third  revolu- 
tions. 

Ans.     —^ — '     -z '     t: beoB. 

3  3.3, 


m 


^ 


/ 


CHAPTEE  VIII. 


DIFFERENTIAL  EQUATIONS  OF  MOTION  OF  A  POINT.* 


Free  Motion  of  a  Point — Rectangular  Co-ordinates.— Let  a  mov- 
ing point  have  a  position  at  any  instant  given  by  the  co-ordinates  x,  y  and 
z,  and  let  the  distance  described  in  the  interval  of  time  dt  be  ds,  and  let 
the  direction  of  motion  make  the  angles  or,  yS,  y  with  X,  F,  Z.  Then  we 
have 


dx  „      dy  dz 

cos  a  =  —-,    cos  p  =  -^ ,    cos  y  =  t-. 
ds  ds  ds 


The  magnitude  of  the  velocity  is 


V  = 


ds 


(1) 
(3) 


and  its  components  in  the  direction  of  the  axes  are 


ds  dx 

Vx='0  cos  a  =  --  cos  nr  =  — -; 
dt  dt' 

„      ds  ^        dy 


ds 


dz 


(3) 


> 

1 
r  ■ 

» 

'^ 

>Vx 

"6 

■  ■■     .Vz 
X 

y 

/ 

0 

U  " 

■»«  =  t>  COS  V  =  -7-  cos  r  =    ,^ 
dt  dt 

We  have  Vx  positive  towards  the  right,  negative  towards  the  left;  %  posi- 
tive upwards  and  negative  downwards ;  Vz  positive  in  the  direction  OZ, 
negative  in  the  opposite  direction  (page  44). 

Squaring  equations  (3)  and  adding,  since  cos"  a  +  cos'  yS  -1-  cos'  ;k  =  1, 
we  have 

.  =  ...v.v  =  (§)-=(|)V(f)%(D-,..(.) 


or 


dt      ^  \dt) 


+  l^J  -^  (l)° 


dy 


(5) 


Let  the  acceleration  be  /  and  its  components  in  the  direction  of  the 
axes  X,  F,  ZhQfx,fy,fz,  then  we  have 


=  ^.     f.= 


d'z 


(6) 


*  This  Chapter  must  be  omitted  by  those  not  familiar  with  the  Calculus. 

81 


82  KINEMATICS— GENERAL  PRINCIPLES.         [CHAP.  VIII. 

The  acceleration  /  is  then 


We  have  fx  positive  towards  the  right,  negative  towards  the  left;  fy 
positive  upwards,  negative  downwards;  fz  positive  in  the  direction  OZy 
negative  in  the  opposite  direction. 

The  tangential  acceleration  ft  =  a  is,  the  rate  of  change  of  speed,  or 

„              dv      (Ps  ^„. 

^'  =  ''  =  M=dt^ (^> 

Differentiating  (4)  and  substituting  (6),  we  have 

vdv  =fxdx  +  fydy  +  fzdz. 

Hence 

r  ^     .     .     .     (9) 

tr*  =  2  /  ifxdx  +  fydy  +  fzdz)  +  Const. 

cLs  d^s 

Dividing  by  ds,  since  v  =  -^,  and  dv  =  -— - ,  we  have  from  (8) 

„  dv       cPs       ^clin        „  dy  dz 

dx      d'x     dy     dj^y      dz      d?z  ,^„. 

d»      df      ds      dt^       da      dC  ^  ^ 

The  normal  acceleration  /„,  as  we  have  seen  (page  76),  is  /„  =  — ,  where 

P 
p  is  the  radius  of  curvature  at  the  point.     We  have  from   analytical- 
Geometry 

p"       \d»^  j        \ds*  J        \as  / 
Hence 

and  the  acceleration/  is  

f=Vfi'+fn'. (11&) 

If  we  denote  by  S  the  angle  which  the  acceleration  /  makes  with  the 
radius  of  curvature  p,  and  by  e  the  angle  which  it  makes  with  the  tangent 
to  the  curve,  we  have  S  +  e,=  90°  and 


tan  e  =  ^  =  -^'- (12) 


'''m 


The  moment  of  the  velocity  about  the  origin  is  the  sum  of  the  mo- 
ments of  the  components.  The  moments  in  planes  parallel  to  XT",  YZ, 
ZXaxQ : 


about  s,    Mz=  —■  X  —  —  -y; 
dt  dt 

.,  ,.        <7s  dy 

"     X,     Mx=  —  ■  V  —  —  •  z: 
"^       dt    "       dt       ' 


dx  dz 

dt  '^~  di 


My  =  -rr  ■  e  —  —.  ■  X. 


(13a). 


CHAP.  VIII.]    DIFFERENTIAL  EQUATIONS  OF  MOTION  OF  A  POINT.     83 

The  moment  about  the  origin  of  the  resultant  velocity  t?  if  j?  is  its 
lever-arm  is  then 

vp  =  M=  i/Mx""  +  My"  +  mF. (186) 

The  line  representative  of  this  moment  makes  the  angles  d,  e,  f  with 
the  axes  of  JT,  F,  Z  given  by 

J         Mx  My  ^         Mz 

cosd  =  — ,   cose  =  -^,  cos/ =-- (14) 

Looking  along  this  line  representative  towards  the  origin,  the  direction 
of  rotation  is  always  counter-clockwise. 

In  the  same  way  the  moment  of  the  acceleration  about  the  origin  is 

the  sum  of  the  moments  of  the  components.    We  have  then  precisely  the 

d?x     d?v     d?z 
same  equations  as    (13),   (14),  only  we  put    — -,    — f-,    ■—  in  place  of 

djf       0/L       aXf 

(lOR        O/U        o& 

-r-y    -^,    3-  in  order  to  find  the  moments  in  the  co-ordinate  planes. 
at      at      at 

Application  of  the  Preceding  Formulas.— Equations  (3)  and  (6)  are 

the  general  equations  by  which  the  motion  of  a  point  is  determined. 
Applications  of  the  use  of  the  equations  just  deduced  will  be  given  here- 
after.    We  can  only  indicate  here  the  general  application. 

If  ^  =  0,  we  have  motion  of  a  point  in  a  plane  only.  The  correspond- 
ing equations  are  at  once  obtained  by  making  z  and  dz  zero  wherever  they 
occur  in  the  general  equations. 

If  we  also  make  /3  =  0  and  ^  =  0,  we  have  motion  along  the  axis  of  x 
only.     Hence  taking  x  =  s,  we  have  from  (3) 

v=-;    from  (6), /  =  /,  =  «  =  —; 

which  are  equations  (8),  (9)  of  page  51. 

If  the  velocity  v  in  any  case  is  given,  it  can  be  resolved  by  (3)  into  its 
components  Vx,  Vy ,  %.  Then  by  differentiating  as  indicated  by  (6)  the 
components /x,fy,  fz  of  the  acceleration / can  be  found,  and  the  accelera- 
tion /  can  then  be  found  by  (7). 

If  the  component  accelerations  are  given,  we  find  by  integration  the 
component  velocities  and  then  the  resultant  velocity. 

If  the  path  is  required,  each  of  equations  (6)  must  be  integrated  twice, 
thus  introducing  two  constants  of  integration  for  each.  The  constants 
of  the  first  integration  will  depend  on  the  initial  velocity,  those  of  the 
second  on  the  initial  position.  We  thus  obtain  equations  involving  x,  y,  z 
and  t,  and  by  eliminating  t  we  obtain  an  equation  between  x  and  y,  or  y 
and  ^,  or  z  and  a;,  that  is,  the  equation  of  the  projection  of  the  path  ou 
the  co-ordinate  planes. 

Differential  Polar  Equations  for  Motion  of  a  Point  in  a  Plane. — 
Let  X  and  y  be  the  rectangular  co-ordinates,  and  ^_^ 

r  and  9  the  polar  co-ordinates,  of  a  point  P  in  a  /   ""^-.^ 

plane.     Then  P/  " 

ic  =  r  cos  6,     y  =  r  sin  9 (15) 

Differentiating  and  dividing  by  dt,  we  have 

dx      dr        .  •    r,dQ  •  ,^-, 

«?«=--  = —-cose  — rsme —: (16) 

dt       dt  dt' 

dy      dr   .    .  .dQ  .^„. 

%  =  -^  = -— smfl -hrcose— - (17) 

^      dt       dt  dt  ^    ' 


84  KINEMATICS— GENERAL  PBINCIPLES.         [CHAP.  VIII. 

Squaring  and  adding,  since  sin'  6  +  cos"  6  =  1,  we  have  for  the  magni- 
tude of  the  velocity 

-(§)■=  (0-'(f)' (-> 

If  r  is  constant,  the  path  is  a  circle.     In  this  case  — -  is  zero  and  v  = 

at 

r——=  roo,  where  oo  is  the  angular  speed  (page  76). 

The  velocity  along  the  radius  vector  is 

-TT  = -TT  cos  e  +  -/- sin  9 (19) 

dt       dt  dt 

The  velocity  perpendicular  to  the  radius  vector  is 

i"-^  =  -jf  cos  fl —  sin  9 (20) 

dt       dt  dt 

fj  or  fl^'}i 

Since  by  (6)  -—  and  -~  are  the  horizontal  and  vertical  components 

of  the  acceleration,  we  have,  by  differentiating  (16)  and  (17), 

^^=w=\dt^-i-dt)  \''''-['di-dt'-'dPi''''''^  ^'i> 
•^^=d^=\dF-idFi  r^'-'rczF^+'^J'''''  ^''^ 

The  acceleration  along  the  radius  vector  is  then  fx  cos  ^  +fy  sin  9,  or 

''''^d^''''=d^-idi) ^^^> 

If  r  is  constant,  the  path  is  a  circle.    In  this  case  -—  is  zero,  and  the 

acceleration  along  the  radius  vector  is  /n  =  —  rco^,  where  go  is  the  angular 
speed  (page  76).  The  (— )  sign  denotes  direction  towards  the  centre 
(page  50). 

The  acceleration  ^erpewc^icwZar  to  the  radius  vector  is 

/j,cos9  —/a;  sin  9, 
or 

^cos9  —  TV  sm9  =  2—  —  +  r-— - (24) 

dt^  dt''  dt  dt        dt""  ^    ' 

If  r  is  constant,  the  path  is  a  circle,  and  —  is  zero,  and  the  acceleration 

perpendicular  to  the  radius  vector  is  ft  =  ra,  where  a  is  the  rate  of 
change  of  angular  speed  (page  76). 
Equation  (24)  may  be  written 

^dtdt'-'-df  =  i:diT'di) '^'^ 

From  equation  (13a)  we  have,  by  inserting  the  values  of  x,  y  and  ^-, 

dt 


d'y 
dt 


CHAP.  VIII.]    DIFFERENTIAL  EQUATIGlJrS  OF  MOTION  OF  A  POINT.      85 

dy 

-^  from  (16),  (17),  for  the  moment  of  the  velocity  with  reference  to  the 

pole,  if  p  is  the  lever-arm, 

vp  =  M^=r'-=r'co, (36) 

where  oo  is  the  angular  speed  (page  76). 

From  equations  (14)  and  (21)  and  (22)  we  have  in  like  manner,  for  the 
moment  of  the  acceleration, 

„         l^dr  de        <fe\ 

^^=i'dfdT'-'de) <2^> 

We  see  from  (25)  that  this  may  also  be  written 

d(rm 

fp=:A^=f^^  =  r'a, (28) 

where  ft  is  the  tangential  acceleration  and  a  is  the  rate  of  change  of 
angular  speed  (page  76). 

Applications  of  the  use  of  these  formulas  will  be  made  hereafter. 

General  Polar  Equations  of  Motion  of  a  Point  in  a  Plane — Ac- 
celeration Central. — When  the  acceleration  is  always  directed  to  or  from 
a  fixed  point  it  is  called  central  acceleration,  and  the  fixed  point  is 
called  the  centre  of  acceleration.     Let  this  fixed  point  be  the  pole. 

Then,  since  the  direction  of  the  acceleration  always  passes  through  the 
pole,  its  moment  with  reference  to  the  pole  is  zero,  and  we  have  from  (28) 

i"^)='>'  "'"§=' w 

where  c  is  a  constant  of  integration. 

dQ 
Now  -5—  =  oa  =  angular  speed,  and  from  page  75  we  have 
at 

roo  =  v  sin  e, 

wiere    e   is    the   angle    which    the    velocity  p 

at  any  point  makes  with  the  radius  vector.  — r" 

Therefore  '^ 

r" —  =  r'oo  =  rv  sm  e  —  e.  5-7^ — ' —   P 

dt  f^\°  I 

From  page  76  we  see  that  r^ta  is  the  mo-  \     g   | 

ment  of  the  velocity  and  is  equal  to  twice  the  ^X" 

areal  velocity  of  the  radius  vector.  \T 

Hence  in  central  acceleration,  the  moment  of 
the  velocity  about  the  pole  is  constant,  the  area  described  by  the  radius 
vector  in  a  unit  of  time  is  constant,  and  tJie  radius  vector  tJierefore  de- 
scribes equal  areas  in  equal  times. 

The  constant  c  is  twice  the  area  described  by  the  radius  vector  in  a  unit 
of  time  (page  61). 

If  p  is  the  perpendicular  let  fall  from  the  pole  upon  the  direction  of  the 
velocity,  we  have 

vp  =  r" —  =  r'fij  =  rv&m  €  =  e (30) 

^         dt 

From  (30)  we  have 

do  c  ,  c  . 

—  —00=:.—      and     v  =  — (31) 

dt  r  P 


86  KINEMATICS— GENERAL   PRINCIPLES.         [CHAP,  VIII. 

Hence  for  central  acceleration  the  angular  speed  at  any  point  of  the 
path  is  inversely  as  the  square  of  the  radius  vector,  and  the  linear  velocity 
at  that  point  is  inversely  as  the  perpendicular  distance  from  the  centre  of 
acceleration  to  the  tangent  to  the  path  at  that  point. 

If  /  is  the  central  acceleration  along  OP,  then  the  component  of  /  nor- 

toial  to  v  is/n=/sin  e,  or,  since  from  (30)  sin  e  =  —fn  =/— .     But  from 

r  r 

v^ 
page  53  we  have  seen  that  fn  =  — ,  where  /a  is  the  radius  of  curvature. 

P 
Hence 

/^  =  ^,    or    v^  =  2fxU2pA 
r       f}  4:\     r  J 

But  2p  sin  e  =  2p—  is  the  length  of  the  chord  of  curvature  2FB  through 

the  pole.     (See  figure  page  85.) 

From  page  28  we  have  for  a  point  moving  from  rest  with  uniform  rate 
of  cliauge  of  speed  a,  if  =  2a(s  —  Si).  Therefore,  for  central  accelera- 
tion the  speed  at  any  point  of  the  path  is  equal  to  that  acquired  by  a 
point  moving  from  rest  with  constant  rate  of  change  of  speed  f  through  a 
space  equal  to  one  fourth  the  chord  of  curvature  through  the  centre  of 
acceleration. 

If  the  acceleration  is  central,  its  component  perpendicular  to  the  radius 
vector  is  zero,  since  the  pole  is  the  centre  of  acceleration,  and  we  have 
from  (24) 

4I-I^=« <-) 

The  component  along  the  radius  vector  is  equal  to  the  acceleration 
itself,  if  the  pole  is  the  centre  of  acceleration,  and  we  have  from  equation 
(23),  if  the  acceleration  is  towards  the  centre, 

cPr        fdQV  .  „„. 

where  the  (— )  sign  for /denotes  motion  towards  the  centre  (page  50). 

Equations  (32)  and  (33)  express  all  the  conditions  of  central  accelera- 
tion towards  the  pole,  and  therefore  determine  the  motion. 

If  the  acceleration  is  away  from  the  pole  we  have  +  /  instead  of  —  / 
(page  50). 

From  (30)  we  have 

d^_^ 
df  ~r*' 
and  equation  (33)  becomes 

^=^-/=---/- w 

But  we  have  seen,  page  76,  that  roo'  is  the  central  acceleration  for  a 
point  moving  in  a  circle  of  radius  r  with  the  speed  roo. 

d?r 
The  rate  of  change  of  length  of  the  radius  vector  ^,  we  see  from  (34), 

is  then  the  difference  between  the  central  acceleration /at  any  instant  and 
the  central  acceleration  at  the  same  instant  of  a  point  moving  in  a  circle 
•of  radius  r  with  the  same  angular  velocity. 

This  rate  of  change  of  velocity  along  the  radius  vector  is  called  the 


CHAP.  VIII.J    DIFFERENTIAL  EQUATIONS  OF  MOTIONOF  A  POINT.      87 

dr 
paracentric  acceleration.  Its  integral  or  t^  is  the  velocity  of  approach  or 

recession  along  the  radius  vector  and  is  called  the  paracentric  velocity. 
(a)  To  find  the  speed  at  any  point  of  the  path. — Central  acceleration. 
If  we  multiply  (32)  by  rdB  and  (33)  by  dr  and  add,  we  have 

or 

Integrating,  we  obtain 

— dr-  =  ''-^J^^^^ (^«> 

■where  Ci  is  a  constant  of  integration. 

If  the  law  of  variation  of  /  is  given  in  terms  of  r  for  any  given  case, 

we  can  perform  the  integration  denoted  by    /  fdr. 

From  (18)  we  have 

._d^_  dr"  +  y-'de' 
~  dt^~         df 
Hence 

^  =  c,  -  2  Cfdr (37) 

"We  have  also  from  (30) 

v  =  - (38) 

*  P 

Since  in  (37)  the  value  of  v  depends  only  upon  r,  we  see  that  the  speed 
J'or  central  acceleration  at  any  two  points  of  the  path  is  independent  of 
thepath^  and  is  the  same  for  any  points  equally  distant  from  the  centre, 
the  law  of  acceleration  remaining  the  same. 

{b)  To  find  the  time  of  describing  any  portion  of  the  path.— Central  accelera- 
tion. 

Substituting  (31)  in  (36),  we  have 

%^^  =  '>,-iffdr. (39) 

Hence 

dr 

We  have  also  from  (31) 

t  =  1  pi^dQ, (41) 

cje, 

from  which  r  must  be  eliminated  by  means  of  the  equation  of  the  path  and 
the  integration  performed  in  reference  to  9. 

(c)  To  find  the  equation  of  the  path. 

r*dO'* 

Substitute  in  (39)  for  dt^  its  value  from  (31),  df"  =  -^,  and  we  have 


,=  r        ";_ ,40) 


i^*h)='-'P''^ <^ 


88  KIKEMATICS — GENEEAL  PRINCIPLES.         [CHAP.  VIII. 

This  equation  may  be  simplified  by  putting  r  =  - -,  and  it  then  becomes 

(cf)  To  find  the  law  of  the  acceleration  for  any  given  path. — Central  accelera- 
tion. 

Differentiating  (43)  With  reference  to  dB,  we  have 

•^=M'^'^'') ^^^^ 

r*dB^ 
Substituting  in  (33)  for  dt"^  its  value  from  (31),  df^  =  —^y  we  have 

^=?[r-^^l' ^''> 

which  Is  the  same  as  (44)  if  we  put  —  =  u. 

r 
T*dft^ 
From  (30)  we  have  p"  =  -^,   and  from    (18)  i?d^  =  (f r*  +  r'dS'. 

Therefore 

di^  +  r^dS' 
Substitute  this  in  (42)  and  we  have 

^  =  c,  - '^i  J fdr. (47) 

Differentiating  with  respect  to  p, 

/=^.    .   .    : (48) 

The  law  of  acceleration  is  given  by  (44)  or  (45)  or  (48). 
From  (33)  and  (35)  we  have 

^,g^_2_dr^ (49> 

r 
an  expression  which  will  often  be  found  useful  in  reductions. 

Differential  Equations  for  Constrained  Motion  of  a  Point  in  a 
Plane. — For  free  motion  of  a  point  in  a  plane  we  have  from  (6)  for  the 
horizontal  and  vertical  components  of  the  acceleration 

Under  the  action  of  these  components,  the  point,  if  free  to  move,  de- 
scribes some  curve. 

But  if  it  is  constrained  to  move  in  a  given  curve,  these  components 
will  be  changed  by  reason  of  the  normal  acceleration  N  due  to  the  given 
curve. 

If  thus /is  the  acceleration  of  a  free  point  P,  fx  and /j,  its  horizontal 
and  vertical  components,  N  the  normal  acceleration  due  to   the  given 
V  f  curve,  and  6  the  angle  of  the  tangent  at  P 

u    {y    I  with  the  horizontal,  we  have 

4 
.U  g-=/.-Mne;      ..    .     .     (50) 

^  ^=/i/  +  iVcose.     .   .   .    (51) 


CHAP.  VIII.J    DIFFERENTIAL  EQUATIONS  OF  MOTION  OF  A  POINT.     89 

If  iV  is  zero,  the  motion  is  unconstrained  and  we  have  (6).  These  two 
equations,  together  with  the  equation  of  the  given  curve,  are  sufficient  to 
determine  the  motion  completely. 

In  applying  them, /a;  is  positive  towards  right,  negative  towards  left, 
and  the  horizontal  component  of  iV  follows  the  same  rule.  We  heiwe  fy 
positive  upwards  and  negative  downwards,  and  the  vertical  component  of 
JV  follows  the  same  rule. 

If  we  multiply  (50)  by  2dx  and  (51)  by  2dy  and  add,  we  have,  since 

sin  G  =  —  and  cos  6  =  ^-, 
as  OS 

?^^^^^^^=2(/.efc.+/,rfy). 
The  first  member  of  this  equation  is  the  differential  of -jfi^  =^ 

•—  =  «",  or  is  equal  to  'iivdv.     Hence 

vdv  =:fxdx  +fydy,  '  1 

or  !■•    •    •    (^2) 

«'      -      -  -  -        -  -  - 


»'  =  3  fWxdx  +fydy)  +  Constant.  J 


This  is  precisely  the  same  result  as  that  obtained  for  free  motion,  equa- 

Hence  we  conclude  that  if  there  is  no  acceleration  except  that  of  iVdue 
to  the  curve  alone,  or  if /x  =  0,  /j,  =  0,  the  speed  on  the  curve  is  constant 
and  unaffected  by  the  curve. 

If  there  is  an  acceleration  besides  that  due  to  the  curve,  the  speed  will 
be  unaffected  by  the  curve  and  the  same  as  if  the  point  were  free. 

If  the  acceleration  of  the  free  point  is  parallel  to  the  axis  of  y,  we  have 
/x=0  and 

««  =  2  ffydy  +  Constant (53) 

If  in  this  last  case/y  is  constant,  we  have 

««  =2fyy  +  Constant (54) 

If  the  distance  of  the  point  from  the  origin  y  =  Si  when  «  =  t?i ,  we 

have                                                                    ^  ,__. 

«"=«,»  +2/j,  (2/- «.), (55) 

which  is  precisely  the  same  as  for  uniform  rate  of  change  of  speed  for  a 
free  point,  as  given  by  eq.  (7),  page  56.  ,  .        ,  ... 

Regard  must  be  had  to  the  signs  in  applying  these  equations  to  any 
special  case.  Velocity  and  acceleration  upwards  are  positive,  downwards 
negative ;  to  the  right  positive,  to  the  left  negative. 

(a)  To  find  the  time  of  motion  of  a  point  on  a  given  curve. 

In  all  cases  rf<  =  — .     Hence  when  the  nature  of  the  curve  and  the 

V 

speed  at  any  point  of  it  are  known,  the  value  of  v  may  be  found  from  (52) 
and  substituted,  and  then  t  may  be  found  by  integration 

If  the  acceleration  of  the  point  is  constant  and  equal  to /and  parallel 
to  y,  we  have  from  (55) 

dt=     ,  ^  (56) 

4/®l'    +   2/y(y-8l) 


^0  KINEMATICS — GENERAL   PRINCIPLES.         [CHAP.  VIII. 

(b)  To  find  the  normal  acceleration  due  to  the  carve. 
If  we  multiply  (50)  by  ~  and  (51)  by  j-  and  subtract,  we  have,  since 

sin e  =  -^,  cose  =  -;-,  and  dx^  +  dy^  =  ds'', 
ds  ds 

''''ds      ''^ds  dsdf 

ds 
Eliminating  dt  by  the  equation  «  =  — ,  we  have 

dt 

dy        r  dx  dxd-'y  -  dyd"" 


But  if  p  is  the  radius  of  curvature  of  the  constraining  curve  at  the 
point  X,  y, 

d^ 

dxd^y  —  dyd^x  ' 
Hence 

ds  ds       p 

=/a, sin e —/j, cose  H ,    (57) 

The  first  two  terms  give  the  normal  component  of  /  for  free  motion. 
The  last  term  is  the  normal  acceleration  due  to  the  curve. 

If  fx  and  fy  are  zero,  the  only  acceleration  is  that  due  to  the  curve  and 

N=~  (page  76). 


KINEMATICS   OF  A   POINT. 

TRANSLATION. 


CHAPTEE  I. 


RECTILINEAR  MOTION  OF  TRANSLATION. 

FAIililNG      BODY,        ACCELERATION      INVERSELY     AS      THE      SQUARE      OP      THE 

DISTANCE. 

Translation. — We  have  defined  translation  (page  13)  as  motion 
of  a  rigid  system,  such  that  every  straight  line  joining  any  two 
points  remains  always  parallel  to  itself.  The  paths  of  all  the  points 
are  therefore  parallel  at  every  instant  and  equal  for  any  given 
interval  of  time,  and  the  velocities  of  all  the  points  at  any  instant 
are  equal  and  parallel. 

If  these  velocities  are  uniform,  that  is,  if  all  points  move  in 
parallel  straight  lines  with  equal  speed,  the  translation  is  uniform. 
If  these  velocities  change  either  in  magnitude  or  direction,  the 
translation  is  variable. 

When,  then,  a  body  has  motion  of  translation  only,  the  motion 
of  the  body  is  the  same  as  that  of  any  one  of  its  points,  and  the 
study  of  the  kinematics  of  a  point  is  therefore  the  study  of  the 
translation  of  a  body. 

Rectilinear  Motion. — If  the  direction  of  the  acceleration  of  a 
point  does  not  change  and  always  coincides  with  the  direction  of 
the  velocity,  then  the  velocity  may  change  in  magnitude  but  can- 
not change  in  direction,  and  we  have  motion  in  a  straight  line. 

In  such  case  the  magnitude  of  the  velocity  is  the  speed  in  a 
straight  line,  the  magnitude  of  the  acceleration  is  the  rate  of  change 
of  speed  and  may  be  either  uniform  or  variable,  and  the  equations 
of  pages  28  or  51  apply. 

Acceleration  is  Proportional  to  Force. — Although  we  are  now 
studying  change  of  motion  without  reference  to  its  cause,  it  will  be 
well  for  the  student  to  keep  in  mind  the  fact  that  no  material  body 
can  change  its  own  motion.  Any  change  of  motion  is  always  found 
to  be  due  to  the  action  of  other  bodies.  This  action  of  external 
bodies  upon  the  body  considered  to  which  change  of  motion  or  ac- 
■celeration  is  due  is  called  force. 

91 


92  KINEMATICS   OF   A   POINT — TRANSLATION.         [CHAP.  I» 

The  student  may  figure  to  himself  such  a  force  as  the  pressure 
or  pull  of  an  imponderable  spiral  spring  upon  the  body,  the  axis  of 
the  spring  having  always  the  direction  of  the  acceleration,  and  the 
spring  moving  with  the  body,  so  that  its  pressure  or  pull  is  exerted 
during  the  entire  time  of  acceleration  and  is  always  proportional 
to  the  acceleration. 

If  the  acceleration  changes  in  direction,  the  axis  of  this  spring 
changes,  so  that  it  is  always  in  the  same  direction  as  the  accelera- 
tion. 

If  the  acceleration  changes  in  magnitude,  the  pull  or  push  of  the 
spring  changes  correspondingly. 

If  the  acceleration  is  uniform,  that  is,  does  not  change  either  in 
direction  or  magnitude,  the  axis  of  the  spring  does  not  change  in 
direction  and  its  pull  or  push  is  constant. 

The  force  of  gravity  upon  bodies  near  the  surface  of  the  earth  is 
like  the  action  of  such  a  spring.  Its  action  is  practically  constant 
in  intensity  and  direction. 

The  student  should  note  that  the  direction  of  the  force  or  accel- 
eration is  not  necessarily  that  of  the  motion,  except  in  the  case  of 
rectilinear  m.otion. 

Thus  in  the  case  of  a  point  moving  with  uniform  speed  in  a 
circle,  the  direction  of  motion  at  any  instant  is  tangent  to  the 
circle,  but  the  acceleration  is  always  directed  towards  the  centre. 

Central  Acceleration. — When  the  acceleration  is  thus  always 
directed  towards  or  away  from  a  fixed  point,  it  is  called  central  ac- 
celeration, and  the  fixed  point  is  called  the  centre  of  acceleration. 

If  the  direction  of  the  acceleration  is  towards  the  centre,  the  ac- 
celeration is  negative  (page  50)  and  the  force  attractive.  If  away, 
it  is  positive  and  the  force  repulsive. 

Uniform  Acceleration — Motion  Rectilinear — Force  Attractive. — 
When  the  direction  of  the  uniform  acceleration  coincides  with  that 
of  the  motion,  we  have  motion  in  a  straight  line  with  uniform  rate 
of  change  of  speed,  and  equations  (2)  to  (7),  page  28  or  51,  apply. 

The  most  common  instance  of  such  motion  is  that  of  a  body  fall- 
ing freely  near  the  earth's  surface.*  In  this  case  the  acceleration 
due  to  gravity  is  known  to  be  practically  constant  and  is  always 
denoted  by  g.  We  have  then  simply  to  replace  a  or  /  by  gr  in  equa- 
tions (2)  to  (7),  page  28  or  51.  We  shall  take  g  =  32.2  ft.-per-sec. 
per  sec.  or  981  cm.-per-sec.  per  sec.  unless  otherwise  specified. 

Value  of  g. — The  value  of  g  is  usually  given  in  feet-per-sec.  per 
sec.  or  in  centimeters-per-sec.  per  sec. 

It  has  been  determined  by  much  careful  experiment  and  found 
to  vary  with  the  latitude  A  and  the  height  h  above  sea-level. 

*  Strictly  speaking  there  is  no  known  instance  in  nature  of  a  uniform  ac- 
celeration (or  of  a  force  wliich  does  not  vary  in  magnitude  and  direction).  The 
acceleration  g  due  to  gravity  (or  the  force  of  gravity)  varies  inversely  as  the 
square  of  the  distance  from  the  centre  of  the  earth  for  a  body  outside  tin; 
earth,  and  directly  as  the  distance  for  a  body  inside,  i.e.,  in  a  shaft  or  well. 

But,  as  we  shall  see,  the  variation  due  "to  this  cause  is  insensible  for  all 
ordinary  distances.  The  decrease  of  ^  at  a  distance  of  a  mile  above  the  earth's 
surface  is  only  about  the  2000th  part  of  its  value  at  the  surface.  Also  two 
radii  of  the  earth  are  sensibly  parallel  when  near  together.  It  is  therefore 
customary  and  practically  correct  to  speak  of  ^r  as  a  constant  acceleration  at  any 
place. 

It  should  be  borne  in  mind,  however,  that  even  then  the  resistance  of  the 
air  very  materially  modifies  the  results  for  falling  bodies.  We  can  therefore 
only  assume  g  as  constant  for  fall  in  vacuo. 


CHAP.  I.] 


RECTILINEAR  TRANSLATION". 


93 


The  general  value  is  given  by 

g  =  32.173  -  0.0821  cos  2A.  -  0.000003^, 

where  h  is  the  height  above  sea-level  in  feet,  and  g  is  given  in  feet- 
per-sec.  per  sec,  or 

g  =  980.0056  -  2.5028  cos  2A  -  0.000003^, 

where  h  is  the  height  above  sea-level  in  centimeters  and  g  is  given 
in  centimeters-per-sec.  per  sec. 

It  will  be  seen  that  the  value  of  g  increases  with  the  latitude,  and 
is  greatest  at  the  poles  and  least  at  the  equator.  It  also  decreases  as 
the  height  above  sea-level  increases. 

The  following  table  gives  the  value  of  g  at  sea-level  in  a  few 
localities : 

9  9 

Latitude.       F.  S.  Units.      C.  S 

Equator 

New  Haven 

Latitude  45° 

Paris 

London 

Greenwich 

Berlin 

Edinburgh 55  57 

Pole 

United  States ] 

For  calculations  where  great  accuracy  is  not  required  it  is  cus- 
tomary to  take  gf  =  32  ft.-per-sec.  per  sec.  or  g  =  981  cm.-per-sec. 
per  sec. 

For  the  United  States  g  =  32^  is  a  good  average  value  and  is 
therefore  very  often  used. 

In  exact  calculations  the  value  of  gfor  the  place  must  be  used. 

Formulas  for  a  Body  Projected  Vertically  Up  or  Down.  —  We 
have  then,  for  a  body  projected  vertically  upwards  in  vacuo, 
simply  to  put  —gia.  place  of  /  in  equations  (2)  to  (7),  page  51.  We 
thus  obtain  ' 


Latitude. 

F.  S.  Units. 

C.  S.  Units. 

0°0' 

32.091 

978.10 

4118 

32.162 

980.284 

45    0 

32.173 

980.61 

48  50 

32.183 

980.94 

5140 

32.182 

980.889 

5129 

32.191 

981.17 

52  30 

32.194 

981.25 

55  57 

32.203 

981.54 

90   0 

32.255 

983.11 

U9   0 

32.162 

980.26 

25   0 

32.12 

979.00 

v  —  Vi  —  gt; 

Vi  —  V 


t  = 


+Vi' 


f-^ 


V  +  Vi,  .        1 

S  —  Si  =        g       t  =  Vit  —  -i 


(1) 

(2) 
(3) 


S  —  Si 


.  _  2(s  -  si)  _Vi  ±  Vvi''  -  2flr(g  -  si) .    ^  ^^^ 
~   V  +  Vi                          g 

v"  =  Vi»  -  2g(s  -  Si) ; (5) 

^'"  - «' .  (6) 


2g 


If  the  starting-point  is  below  the  origin,  we  should  change  the 

sign  of  Si.  ,         ,  -.      ,  .Li-         • 

If  the  body  is  projected  downwards,  we  should  change  the  signs 
of  v,vi,s  and  Si.    We  see  that  this  is  equivalent  to  simply  chang- 


/I 


94    KINEMATICS  OF  A  POINT — KECTILINEAR  TBANSLATION.  [CHAP.  T, 

ing  the  sign  of  g  in  all  equations,  leaving  the  signs  of  the  other 
quantities  unchanged. 

When  the  final  velocity  v  is  zero,  we  have  from  (2),  for  the  time 
of  rising  to  the  highest  point  or  the  "  turning-point,'''' 

T  =  — 

9  ' 

For  the  time  of  rising  to  the  highest  point  and  returning  to  the 
starting-point  we  make  s  —  Si  =  0  in  (4)  and  obtain 

2T  =  ^. 
9 

Hence,  the  times  of  rising  and  returning  are  equal. 

For  body  falling  we  have  T  = \  the  minus  sign  denoting 

time  before  the  start  necessary  to  acquire  the  velocity  Vi. 

The   distance  from  the  starting-point  to  the  turning-point  is 

found  from  ( 6 j,  by  making  t?  =  0,  to  be  ^. 

2g 

The  distance  ^  or  -  is  called  the  height  due  to  the  velocity  Vi  or 

V ;  that  is,  the  distance  a  body  must  fall  from  rest  in  order  to  ac- 
quire the  velocity  Vi  or  t?. 

When  the  distances  in  rising  and  falling  are  equal  we  have 

s  —  Si  =  0,  or  x^  =  jr— ,  or  tJi  =  V;  that  is,  the  velocity  of  return  is 
2g        2g 

equal  to  the  velocity  of  projection. 

If  the  time  of  rising  is  less  than  T  =  — ,  the  displacement  s  —  Si 
is  equal  to  the  distance  described.  But  if  the  time  of  rising  is 
greater  than  2'  =  — ,  the  body  reaches  the  turning-point  and  then 
falls  from  rest,  and  the  entire  distance  described  is 

distance  described  =  ~  +  -qCt  —  T)^  =  — s  =  -^—- •      (7) 

2g^2^^         '        g  2g 

[Application  of  Calculas  to  the  preceding  Case.] — We  can  deduce  the  pre- 
ceding equations  from  our  general  equations  (8)  to  (10),  page  51. 

Thus  from  equation  (9)  we  have,  for  acceleration  directed  downwards  and 
therefore  (— ),  (page  50,) 

^  ~  dt~  df   ~      ^' 
Integrating,  we  have 

«  =  —  =  —  gt-\-  Const. 
CLt 

When  <  =  0,  let  «  =  -f  «i ,  the  (-|-)  sign  denoting  motion  upwards. 
Then  we  have  Const.  =  Vi,  and 

*  =  S  =  ^'-^*'     •     •     • ^^> 


which  is  equation  (1),  page  93. 
Integrating  again, 


s  =  Dii  —  ~gti  ■\-  Const. 


CHAP.  I.]  CALLING   BODY— EXAMPLES.  96- 

Let  8  =  +  Si  when  t  =  0,  tlie  (-|-)  sign  denoting  distance  upwards. 
Then  we  have  Const.  =  s, ,  and 

8-Si  =  Vit--gt\ (3> 

which  is  equation  (3),  page  93. 

From  equations  (1)  and  (3)  we  can  deduce  all  the  others  page  93. 

The  student  should  note  especially  that  these  equations  have  been  deduced 
for  body  projected  upwards  or  Vi  positive. 

If  we  suppose  motion  towards  the  centre  or  downwards,  we  should  have  ©, 
negative. 

Also  if  the  starting-point  is  below  the  origin,  we  should  change  the  sign  of 
Su 

In  all  cases  we  take  g  minus,  as  long  as  the  acceleration  is  directed  down- 
wards. 


EXAMPLES. 

Unless  otherwise  specified  g  =  32.2  ft.-per-sec,  per  sec.  or  981  cm.-per-sec. 
par  sec.     All  bodies  supposed  to  move  in  vacuum. 

.,  V   v(i)  A  point  moves  with  a  uniform  velocity  of  2  ft.  per  sec.    Find 
ij  '  the  distance  from  the  starting-point  at  the  end  of  one  hour. 
"  ^Ans.  7200  ft.     Motion  in  a  straight  line. 

-•  -(2)  Two  trains  have  equal  and  opposite  uniform  velocities  and 
_    each  consists  of  12  cars  of  50  ft.    They  are  observed  to  take  18  sec. 
to  pass.    Find  their  velocities. 
Ans.  22.73  miles  per  hour. 


^ 


Two  points  move  with  uniform  velocities  of  8  and  15  ft.  per 
sec',  in  directions  inclined  90\  At  a  given  instant  their  distance  is 
10  ft.  and  their  relative  velocity  is  inclined  30°  to  the  line  joining 
them.  Find  (a)  their  distance  when  nearest ;  (b)  the  time  after  the 
given  instant  cut  ichich  their  distance  is  least. 

Ans.  (a)  5  ft. ;  (6)  ^  j/S  sec. 

(4)  A  body  is  projected  vertically  upwards  with  a  velocity  of  300 
ft.  per  sec.  Find  (a)  its  velocity  after  2  sec.;  (b)  its  velocity  after 
15  sec;  (c)  the  time  required  for  it  to  reach  its  greatest  height ; 
id)  the  greatest  height  reached  ;  (e)  its  displacement  at  the  end  of 
15  sec;  if)  the  space  traversed  by  it  in  the  first  15  sec ;  {g)  its  dis- 
placement ivhen  its  velocity  is  200  ft.  per  sec.  upwards ;  (h)  the  time 
required  for  it  to  attain  a  displacement  of  S20ft.* 

Ans.  (a)  235.6  ft.  per  sec.  upwards;  (6)  183  ft.  per  sec.  downwards;  (c)  9.3 
sec;  (d)  1397.5  ft.;  (e)  877.5  ft.  upwards;  (/)  1917.5  ft.;  (g)  776.3  ft.  upwards; 
(h)  1.13  sec.  in  ascending,  17.5  sec.  in  descending. 

{,5)  A  ball  is  projected  upivards  from  a  window  half  ivay  up  a 
toicer  117.72  meters  high,  with  a  velocity  of  39.24  m.  per  sec.  Find 
the  time  and  speed  (a)  with  which  it  passes  the  top  of  the  tower 
ascending ;  (b)  the  same  point  descending ;  (c)  reaches  the  foot  of 
the  tower. 

Ans.  (a)  2  sec;  19.62  m.  per  sec;  (ft)  6  sec;  19.62  m.  per  sec; 
(c)  (4  -f  24/7)  sec;  19.62  i^l  m.  per  sec. 

*  If  the  student  will  refer  to  the  Examples,  page  114,  he  will  gain  an  idea, 
of  the  effect  of  the  air  in  modifying  the  motion  of  falling  bodies,  and  will 
better  appreciate  the  delusive  nature  of  all  problems  which  ignore  it. 


96    KINEMATICS  OF  A  POINT — RECTILINEAR  TRANSLATION.  [CHAP.  I. 

^>/L,(6)  A  stone  is  dropped  into  a  well  and  tJie  splash  is  heard  in  3.13 
kec.  If  sound  travels  in  air  with  a  uniform  velocity  of  332  meters 
per  sec,  find  the  depth  of  the  well. 

Ans.  44.1  meters. 

^7)  If  in  the  preceding  example  the  time  until  the  splash  is  heard 
^  and  the  velocity  of  sound  in  air  is  V,  find  the  depth. 


-:d 


Ans.  Depth  =  -\  (Tg-^  V)-  \/V{2Tg-{-  V) 


/ 


pjS)  Show  that  a  body  projected  vertically  upwards  requires  twice 
as'long  a  time  to  return  to  its  initial  position  as  to  reach  the  highest 
point  of  its  path,  and  has  on  returning  to  its  initial  position  a 
speed  equal  to  its  initial  speed. 

(9)  A  stone  projected  vertically  upwards  returns  to  its  initial 
position  in  6  sec.  Find  (a)  its  height  at  the  end  of  the  first  second, 
and  {b)  what  additional  speed  would  have  kept  it  1  sec.  longer  in  the 
air. 

Ans.  {a)  80.5  ft.;  (J)  16.1  ft.  per  sec. 


(10)  A  body  let  fall  near  the  surface  of  a  small  planet  is  found  to 
traverse  204  ft.  between  the  fifth  and  sixth  seconds.  Find  the  ac- 
celeration. 

Ans.  20.4  ft.-per-sec.  per  sec. 

(11)  A  particle  describes  in  the  nth  second  of  its  fall  from  rest  a 
space  equal  to  p  times  the  space  described  in  the  (n  —  l)th  second. 
Find  the  lohole  space  described. 

ga  -  Spy 


Ans. 


8(1  -  py 


(12)  A  body  uniformly  accelerated,  and  starting  ivithout  initial 
velocity,  passes  over  bfeet  in  the  first  p  seconds.  Find  the  time  of 
passing  over  the  next  b  ft. 

Ans/jo(  '^^2'-  l)  sec. 


[3)  A  ball  is  dropped  from  the  top  of  an  elevator  4.905  meters 
high.  Acceleration  of  gravity  is  9.81  meter s-per -sec.  per  sec.  Find 
the  times  in  which  it  will  reach  the  floor  (a)  when  the  elevator  is  at 
rest;  (b)  ivhen  it  is  moving  with  a  uniform  downward  acceleration 
of  9.81  m.-per-sec.  per  sec;  (c)  when  moving  with  a  uniform  down- 
xoard  acceleration  of  4.905  m.-per-sec.  per  sec.  ;  (d)  ivhen  moving  with 
a  uniform  upward  acceleration  o/ 4.905  m.-per-sec.  per  sec. 


Ans.  (a)  1  sec. ;  (i)  oo ;  (c)  |/3^sec. ;  {d)  y  , 


I  sec. 


(14)  If  Si ,  Si  are  the  heights  to  which  a  body  can  be  projected 
ivith  a  given  initial  vertical  velocity  at  two  places  on  the  eartlvs  sur- 
face at  which  the  accelerations  of  falling  bodies  are  gi  and  g^  respec- 
tively, show  that  Sigi  =  Stg-i. 


5)  A  stone  A  is  let  fall  from  the  top  of  a  tower  483  ft.  high. 
Ai  the  same  instant  another  stone  B  is  let  fall  from  a  window  161 
ft.  below  the  top.     How  long  before  A  will  B  reach  the  ground  f 

Ans.  (  1/6  —  2)  V5"sec. 


■CHAP.  I.]  FALLING   BODY — EXAMPLES.  97       ^ 

u<re)  A  hall  falling  from  the  top  of  a  toicer  had  descended  e^feet 
when  another  was  let  fall  at  a  point  b  below  the  top.     Show  that  if 

they  reach  the  ground  together,  tJie  height  of  the  tower  is  -    "^    '  ft. 

4a 

(17)  If  two  bodies  are  projected  vertically  upwards  with  the  same 
initial  velocity  F,  at  an  interval  of  t  sec,  prove  that  they  will  meet 

at  a  height  ^{-^--^. 

(18)  Tivo  stones  are  falling  in  the  same  vertical  line.  Show  that 
if  one  can  overtake  the  other,  it  ivill  do  so  after  the  same  lapse  of 
time,  even  if  gravity  ceases  to  act. 

(19)  Bodies  are  projected  vertically  doivnwards  from  heights  hi , 
hi ,  hi  with  velocities  Vi,  v^,  Va ,  and  all  reach  the  ground  at  the  same 
moment.    Shoiv  that 

hi  —  hi  hi  —  hi  ha  —  h^ 

Vi  —Vi  Vi  —  Va  Vs  —  Ui  ' 

(20)  Two  points  move  in  straight  lines  with  uniform  accelerations. 
Show  that  if  at  any  instant  their  velocities  are  proportional  to  their 
respective  accelerations,  the  path  of  either  relative  to  the  other  will 
be  rectilinear. 

(21)  Upon  the  top  of  a  toicer  200  feet  high  is  placed  a  flag-staff  of 
26  feet ;  a  bullet  is  Let  fall  from  the  top  of  this  flag-staff,  and  at  the 
instant  of  its  passing  the  bottom  of  it  a  stone  is  l^  fall  from  a  win- 
dow 44  feet  from  the  top  of  the  tower.  At  what  distance  from  the 
bottom  of  the  tower  will  the  bullet  overtake  the  stone  f  Show  also 
that  this  distance  is  independent  of  the  value  of  g.  In  what  tim^ 
after  the  dropping  of  the  stone  do  they  meet  f  In  what  time  after 
the  dropping  of  the  bullet  ?  How  far  does  the  stone  fall  before  meet- 
ing f     Take  acceleration  due  to  gravity  32.16  ft.  per-sec.  per  sec. 


r  44' 

Ans.  Distance  =  200  —     44  +  ^— 


4X26 


=  137.385  ft. 


Time  from  falling  of  stone  1.07  sec.  Time  from  falling  of  ballet  2.34  sec. 
The  stone  falls  18.615  ft.     The  bullet  falls  88.615  ft. 

(22)  A  body  falls  a  distance  a  from  rest  when  another  body  is 
let  fall  from  a  distance  a  +  b  below  the  starting-point  of  the  first. 
How  far  tvill  the  latter  body  fall  before  it  is  overtaken  by  the  for- 
mer ?     What  is  the  time  of  fall  of  the  latter  body  f 


--;     time  =  4/  ^ — . 
4a  r    2ag 

(§3)  A  body  is  projected  upward  with  a  velocity  which  would 
take  it  to  the  height  a,  and  at  the  same  instant  a  body  is  let  fall 
from  a  distance  above  the  point  of  projection  of  b.  At  what  dis- 
tance below  the  latter  point  will  the  bodies  meet  f    In  what  time  f 

Ans.  a;  =  -—  ;     time  =  4/  ^ — . 
4a  r    2ag 

(24)  A  body  is  thrown  vertically  upward' with  a  velocity  Vi.  Find 
the  time  at  which  it  is  at  a  given  height  h  in  its  ascent. 

Ans.    ,  =  ^_L±J^£ZZM. 

a 

The  lower  sign  gives  the  time  when  the  body  is  at  the  height  h  in  ascend- 
ing, the  upper  in  descending. 


98    KIKEMATICS  OF  A  POINT — KECTILINEAR  TRANSLATION^.  [CHAP.  I» 

(25)  A  body  is  projected  vertically  upward  and  the  interval  be- 
tween the  times  of  its  passing  a  point  whose  height  is  h  in  its  ascent 
and  descent  is  2t.  Find  the  velocity  Vi  of  projection  and  the  whole 
time  T  of  motion. 

2\/(ff^  +  2gh 


Ans.  Vi  =    ^gH''  +  2gh,   T  = 


26)  A  body  falling  to  the  ground  is  observed  to  pass  through 
eight  ninths' of  its  original  height  in  the  last  second.      Find  the 
height. 
9 

Ans.  -^  =  36  ft.  nearly. 


'-I^TM 


i7)  A  body  falling  under  the  action  of  gravity  is  observed  to 
describe  144.9  feet  and  177.1  feet  in  two  successive  seconds.  Find  g 
arid  the  time  from  the  beginning  of  the  motion  to  the  first  of  the  two 
seconds. 

Ans.  g  =  32.2  ft.-per-sec.  per  sec,  t  =4  sec. 

(28)  A  falling  body  is  observed  at  one  portion  of  its  path  to  pass 
through  nfeet  in  t  sec.  Find  the  distance  described  in  the  next  t 
seconds. 

Ans.  n  -|-  gf  feet. 

(29)  A  body  is  projected  vertically  upwards  with  a  velocity  '6g. 
At  what  times  wilt  its  height  be  4g,  and  what  will  be  its  velocity  at 
these  times  f 

Ans.  It  will  be  at  the  height  4g  at  the  end  of  2  sec.  and  again  at  the  end  of 
4  sec.  Its  velocity  at  both  these  instants  is  g  ft.  per  sec.  upward  at  the  end  of 
2  sec.  and  downward  at  the  end  of  4  sec. 


(30)  Find  the  velocity  with  which  a  body  must  be  projected  up  a 
smooth  inclined  plane.,  the  height  of  which  is  h  and  length  Z,  to  reach 
the  top. 

Ans.  The  vertical  acceleration  is  g.     The  component  acceleration  parallel 

to  the  plane  is  a  =  ^.     Therefore  «  =  ^2al  —  i^2gh,  or  the  same  as  the  ve- 
locity required  to  project  the  body  to  the  height  h. 

^"'fsi)  Find  the  time  of  falling  down  the  whole  length  of  a  smooth 
inclined  plane  of  length  I  and  height  h. 

Ans    li/ —•    ^^^  constant  height  h  the  time  is  directly  as  the  length. 
r    gh 

^"^2)  Find  the  speed  attained  by  a  body  in  falling  down  a  smooth 
inclined  plane  the  height  of  which  is  h  and  length  I. 

Ans.     \^2gh,  the  same  as  in  falling  through  the  height  7t. 

^•'^tSS)  A  body  is  projected  down  a  smooth  plane  the  inclination  of 
which  with  the  horizontal  is  45°,  with  a  velocity  of  10  ft.  per  sec. 
Find  the  space  described  in  2i  seconds  (g  =  32). 
Ans.  95.7  ft 

^^34)  A  locomotive  starts  down  a  smooth  incline  with  a  velocity  of 

7i  miles  an  hour.    If  the  ratio  of  the  height  to  length  is  gx^,  find  the 

space  traversed  in  two  minutes  {g  =  32). 
Ans.  2472  ft. 


CHAP.  I.]    ACCELERATION  INVERSELY  AS  SQUARE  OF  DISTANCE. 


99 


r  Acceleration  Inversely  as  the  Square  of  the  Distance  from  a  Fixed 
f  Point— Motion  Rectilinear— Force  Attractive.— This  is  the  case  of 
i  a  body  at  a  great  distance  from  the  earth,  under  the  action  of  the 
force  of  gravity,  since  in  such  case  the  acceleration  is  towards  the 
centre  of  the  earth  and  varies  inversely  as  the  square  of  the  dis- 
tance from  the  centre  of  the  earth.* 

Let  si  be  the  initial  distance  from  the  centre  of  acceleration,  the 
velocity  at  this  point  being  v^ ,  and  s  the  distance  to  any  other  posi- 
tion at  which  the  velocity  is  v. 

If  the  acceleration  is  towards  the  centre  or  force  attractive,  it  is 
negative  (page  50). 

Let  a'  be  the  known  acceleration  at  a  distance  r',  and  a  the  accel- 
eration at  any  distance  s.    Then  we  have 


a:a'::r"':  s",     or      a  = 


ar 


Thus  for  instance,  in  the  case  of  the  earth,  a'  is  gr,  and  r'  is  the 
radius  of  the  earth  at  the  locality  where  g  is  known.  -«• 

We  have  then,  if  r  is  an  indefinitely  small  time,  for  acceleration 
towards  the  centre 


=  a  =  — 


(1) 


The  mean  velocity  for  an  indefinitely  short  time  is 


V  +  Vi 


,  and 


the  distance  described  in  this  time  is 


S  —  Si  = 

Multiplying  by  (1), 


-r,    or    V  +  Vi  = 


2(s  —  8i) 


2a'r'\ 

V^  —Vi'= -— (S  —  Si). 

s" 
But  if  the  time  is  indefinitely  small,  s"  will  equal  ssi ,  and  hence 

v^  _  Vi"  =  -  2a'r''(-  -  -V 
\8i        sj 

....     (2) 


or 


v^  =  Vi^  -  2a'r''l-  --\. 


If  the  body  falls,  Si  is  greater  than  s  and  the  last  term  becomes 
essentially  positive.  If  the  body  is  projected  upwards,  Si  is  less 
than  s  and  the  last  term  is  essentially  negative.  Equation  (2)  holds 
good,  then,  without  change  in  either  case  if  the  acceleration  is 
towards  the  centre,  or  force  attractive.    It  also  holds  for  any  path 


*  This  is  the  '  law  of  universal  gravitation  "  as  discovered  by  Newton.  It 
is  also  known  as  the  law  of  the  inverse  squares. 

It  is  regarded  as  rigidly  true  for  every  particle  of  matter  acting  upon  every 
other  particle.  But,  as  we  shall  see,  it  is  not  rigidly  true  for  bodies  of  finite 
dimensions  acting  upon  similar  bodies,  unless  those  bodies  are  homogeneous 
spheres  or  spherical  shells. 

The  earth  is  not  a  sphere  and  is  not  homogeneous.  Therefore  it  is  not 
rigidly  true  that  it  attracts  external  bodies  with  a  force  inversely  as  the  square 
of  the  distance  from  the  centre.  The  deviation  from  this  law  for  bodies  at 
great  distances  is,  however,  insensible. 


100  KINEMATICS  OF  A  POINT — RECTILINEAB  TRANSLATION.  [CHAP.  I. 


straight  or  curved  if  s,  a, ,  r'  are  measured  along  the  path,  and  a'  is 
the  tangential  acceleration  or  rate  of  change  of  speed  at  distance  r\ 
and  V  and  Vi  are  the  speeds  final  and  initial. 

If  the  acceleration  is  away  from  the  centre,  or  force  repulsive, 
we  should  have  the  sign  before  the  last  term  (+)  instead  of  (— ),  or 
the  sign  of  a'  is  changed  in  (1)  and  (2). 

If  the  body  falls  from  rest  from  a  distance  Si ,  we  have  from  (2), 
by  making  Vi  =  0,  for  a  body  falling  from  rest. 


v'  =  2a'r"'(-  -  -]  falling. 


(3) 


If  the  body  is  projected  upwards  and  the  velocity  v  is  zero  at 
the  height  s,  we  have  from  (2),  for  a  body  projected  upwards  to  a 
distance  s, 

Vi'^  =  2aV"( 1  rising (4) 

Cor.  1.  If  the  distance  Si  is  infinite,  we  have  from  (3)  the  velocity 
acquired  in  falling  from  an  infinite  distance, 


v  =  -  y- 


2a'r'^ 


and  from  (4)  the  velocity  of  projection  in  order  to  go  to  an  infinite 
distance  is 


,    ./2a'r" 


vn 


In  the  case  of  a  body  attracted  by  the  earth  we  have  a'  =  g.  If 
then  in  the  first  case  s  =  r'  and  in  the  second  case  Si  =  r',  we  have 
for  the  velocity  acquired  in  falling  to  the  surface  of  the  earth  from 
an  infinite  distance,  or  the  velocity  necessary  to  project  a  body  to 
an  infinite  distance  from  the  surface  of  the  earth,  in  vacuo, 

V  —  Vi  —  T  V2gr'. 

If  we  take  gr  =  32i  ft.-per-sec.  per  sec.  and  the  mean  radius  of 
the  earth  3960  miles,  we  have 

^Q^  X  3960\i      ^  „  ^^      ., 
v  =  Vi  =  [  — ^--^- —  I  =  T  6.95  miles  per  sec. 
\       5280       j 

The  (  — )  sign  for  falling  and  the  (+)  sign  for  upward  projection. 
Cor.  2.  If  we  put  equation  (3)  in  the  form 

V'  =  2gr"l-—  ^ 
\  ss 

we  see  at  once  that  if  Si  —  s  is  a  small  distance  compared  to  s,  so 
that  the  entire  fall  takes  place  near  the  earth's  surface,  ssi  will  be 
practically  equal  to  r"*  and  we  shall  have 

u"  =  2g{Si  —  s). 

This  is  the  same  formula  as  for  uniform  acceleration  g  towards 
the  earth,  the  initial  velocity  being  zero  (page  93). 

[Application  of  Caleolas  to  the  Preceding  Case.] — We  can  deduce  the  preced- 
ing results  from  our  general  equations  (8)  to  (10),  page  51. 

Thus  for  acceleration  towards  the  centre  we  have,  as  before, 

_  dv  _d^s  _      a'r'^  ^. 


dt       dt^ 


Civil  &  Mechanical  Engineer. 

SAN  FRAiN CISCO,  CAL, 

CHAP.  I.]    ACCELERATION  IKVEESELY  AS  SQUARE  OF  DISTANCE.     101 


For  acceleration  away  from  the  centre  we  should  have  (-f-)  instead  of  (— ) 

(page  50). 

ds 
Multiply  by  ds,  both  sides,  and  then,  since  t-  =  ■»,  we  have 


vdv  =  — 


a'r'^ds 


Integrating,  we  have 


<' 


-J«^ 


a'r" 


-f-  Const. 


When  s  =  Si ,  let  v  =  -\-Vi  for  body  projected  upwards  and  v  =  —Vi  for 

body  projected  downwards.     Then  in  both  cases  Const.  =  —-7 ,  and  in 

2  Si 

both  cases 


«»  =  ®,»  -2a'r'^(~  --V 


(2) 


These  are  the  equations  (1)  and  (2)  of  the  preceding  Article.  They  hold,  as 
we  see,  for  motion  towards  or  away  from  the  centre,  provided  the  acceleration 
is  towards  the  centre.  For  acceleration  away  from  the  centre  we  change  the 
sign  of  a'.  These  equations  also  hold  for  any  path,  straight  or  curved,  if  s, 
Si ,  r'  are  measured  along  the  path,  and  a'  is  the  tangential  acceleration  or  rate 
of  change  of  speed  at  distance  /,  and  d  and  Vi  the  speeds  final  and  initial. 

If  the  initial  velocity  is  zero,  we  have  for  a  hodjfaUing  from  rest 


ds 


«»^  2ay''{---\. 


(3) 


Since  ®  =;  — ,  we  have  from  (3) 


-f  =  -A-(^i). 


where  we  take  the  (— )  sign  for  the  radical  to  denote  motion  towa/rds  the  centre   S 
(page  44).     This  can  be  put  in  the  form 

sds  ,^  ,  /%a'r'^ 

—  =  at  4/ . 

4/siS  — s"  ^         *• 

To  put  this  in  a  form  convenient  for  integration,  add  and  subtract  ^«i  to  the 
numerator  of  the  first  term.     We  then  have 


I 


\Sx—S  — 


2« 


Integrating,  we  have 


2  i/s.s  -  8»         2  /« 


SxS  —  «* 


r         Si 


«.*  Si  .  _i  2s 
(siS  —  «*)^  —  TT  versm  — 
^  '        2  « 


s  _  /2a'r'y. 


t  +  Const. 


(4) 


Let  t 


iCSi 


0  when  s  =  Si,  then  Const.  = ~.   Hence*  for  the  time  of  falling 


-,2s                     _i/,       2«\            _,/2s       ,\ 
*  We  have  n  —  versm     —  =  ;r  —  cos    1 1 1  =  cos     ( 11. 

From  trigonometry,  2  cos«  y  —  1  =  cos  2y.   Let  2y  =  cos" '  f  -^ 1  j .     Then 

cos  2v  =  -  —  1  and  cos"  w  =  — ,  or  jr  =  cos^'i/—  and  2y  =  2  cos"'4/  -  . 


\ 


Hence  2  cos 


.    _i2s 
versin      — 

Sl 


102  KINEMATICS  OF  A  POINT — RECTILINEAR  TRANSLATION.  [CHAP.  I. 
[  from  rest  we  have  for  acceleration  towards  the  centre 

;  If  we  had  taken  motion  away  from  the  centre,  we  should  have  obtained, 

'  instead  of  (4), 

^  —  (siS  —  «")  +  ^  versin     —  =  I )  <  -|-  Const, 

.rr  C  8i         \^      8i     J 

Let  t  =  0,  when  8  =  0,  and  Const.  =  0,  and  we  have 

I  Equation  (6)  applies  to  a  body  projected  upwards  to  any  height  s,  Si  being 

the  height  at  which  it  would  come  to  rest. 

If  we  make  s  =  Si  in  (6),  or  «  =  0  in  (5),  we  find  the  time  of  reaching  the 
turning-point  in  rising  or  reaching  the  centre  in  Tailing  from  rest 


AT 


2  ^^'r'^ 


$ 


/ 


CHAPTEE  II. 


SIMPLE  HARMONIC  MOTION.    MOTION  IN  RESISTING 

MEDIUM. 


^  Simple  Harmonic  Motion. — The  motion  of  a  point  moving  in  any 
path  in  such  a  manner  that  the  tangential  acceleration  is  directly 
proportional  to  the  distance,  along  the  path,  from  a  fixed  point  in 
the  path  is  called  simple  harmonic  motion.  Such  motion  may  be 
rectilinear  or  curvilinear. 

The  vibrations  of  such  bodies  as  a  tuning-fork  or  a  piano-wire 
are  approximate  examples  of  such  motion,  and  hence  the  term 
"  harmonic."  The  vibrations  of  an  elastic  body,  such  as  the  air,  are 
examples  of  such  motion. 

It  is  also,  as  has  been  stated  (note,  page  92),  the  motion  of  a 
body  under  the  action  of  gravitation,  within  a  homogeneous 
sphere,  as  it  can  be  shown  that  in  this  case  the  acceleration  due  to 
gravity  is  proportional  to  the  distance  from  the  centre. 

The  motion  of  the  piston  of  a  steam-engine  when  moved  by  a 
crank  and  connecting-rod  approximates  the  same  motion  if  the 
rotation  of  the  crank  is  uniform,  the  approximation  being  closer 
the  longer  the  connecting-rod.  This  wUl  be  evident  from  the  fol- 
lowing Article. 

Simple  Harmonic  Motion  in  a  Straight  Line — Force  Attractive. — 
Let  a  point  M  move  with  uniform  speed  in  a 
circle  of  radius  CM  =  r. 

Then  the  acceleration  fn  is  always  direct- 
ed towards  the  centre  and  equal  to  fn  =  rm", 
where  go  is  the  constant  angular  velocity 
(page  76). 

The  projection  of  /71  upon  the  diameter  A  ^ 
CA  is  fijV  cos  MCP.    But  r  cos  MCP  is  the 

distance  CP=s  of  the  projection  P  of  M  upon  the  diameter  CA. 

Therefore  the  projection  of  fn  upon  the  diameter  is  a  =  m's,  or, 

since   go  is  constant,   a  is  directly  proportional  to  the  distance 

\       CP  =  s.     The  motion  of  P  is  therefore  harmonic. 

\  If  then  a  point  M  moves  with  uniform  speed  in  a  circle,  its  pro- 

^Jection  P  itpon  any  diameter  moves  with  harmonic  motion  in  the 

^__ii/am€fer,  the  centre  of  acceleration  being  the  centre  of  the  circle. 

,  Let  a'  be  the  known  acceleration  in  the  Hne  AC  of  P  at  a  given 

I  /„T 

^^^  i     -distance  r'  from  the  centre.    Then  a'  =  mV  and  00  =a/  ^ ,  and  the 

i  '  * 

103 


104 


i 


\ 


KINEMATICS   OF   A   POIIfT — TRAN8LATI0K.        [CHAP.  II. 


speed  of  M  is  roo  =  r|/?-.    The  projection  of  this  speed  on  the 
diameter  is  4/?^  r  cos  CMP=  MP  a/—. 

'      V  r      -V.' 


But  MP  =  |/r"  —  s'  ;  hence  we  have  for  the  velocity  of  the  point 
P  in  the  lino  ^C  at  the  distance  s  =  CP  from  C, 


=  ^,{r^-8^). 


(1) 


or,  if  Vi  is  the  initial  velocity, 


s\ 


where  a'  is  the  know  n  acceleration  of  P  at  a  given  distance  r'  from 

Thus  the  point  P  starts  from  rest  at  the  distance  s  =  r  from  C. 
The  velocity  increases  as  the  distance  s  decreases,  till  P  arrives 
at  the  centre  C  where  the  velocity  is    a    maximum  and  equal 

to  V  =  r  |/ "-.      Then  the  velocity  decreases  and  finally  becomes 

zero  when  P  arrives  at  A'  at  the  distance  s  =  —  r  on  the  other  side 
of  C.  Equation  (1)  holds  good  for  motion  towards  or  away  from 
the  centre  if  the  acceleration  is  towards  the  centre,  or  force  attract- 
ive. It  also  holds  for  any  path  straight  or  curved  if  s,  r,  r'  are 
measured  along  the  path  and  a'  is  the  tangential  acceleration  or 
rate  of  change  of  speed  at  distance  r',  and  v  and  v'  the  speeds  final 
and  initial. 

Cor.  1.    Since  the  uniform  speed  of  ilf  in  the  circle  is  ri/— ,  the* 

r   ^> 

time  occupied  by  P  in  passing  from  A  to  A'  and  back  to  A  is 


But  if  a  is  the  acceleration  at  any  distance  s,  and  /„  is  the  ac- 
celeration at  the  extreme  distance  r,  we  have  for  harmonic  motion 

r'  _  r  _s  _  JL_ 

a'  ~  fn     a      <»'" 

Hence  the  time  T  of  a  complete  oscillation  is 


GO  'a 


\  a.] 

\ 


The  time  T  of  a  complete  oscillation  depends  therefore  only 
upon  the  constant  ratio  —  =     ,  and  is  independent  of  the  range  r 

or  amplitude  of  the  oscillation.  For  this  reason  the  oscillations  are 
said  to  be  isocbronous,  or  made  in  equal  times,  no  matter  what  the 
range  or  amplitude. 

Cor.  2.  Since  the  motion  of  a  body  under  the  action  of  gravity 
in  a  homogeneous  sphere  is  harmonic  (page  92),  if  we  put  g  for  a' 
and  let  r'  be  the  mean  radius  of  the  earth,  we  have  from  (1)  the 
motion  of  a  body  falling  under  the  action  of  gravity  towards  th& 


CHAP.  II.  J 


SIMPLE    HARMOKIC   MOTIOlf. 


105- 


P 


centre  of  the  earth  in  a  well  or  shaft,  assuming  the  earth  to  be 
a  homogeneous  sphere  and  neglecting  resistance  of  the  air. 
In  such  case  (1)  becomes 

u"  =  %(r  +  s){r  -  8). 
r 

If  the  fall  takes  place  for  a  short  distance  compared  to  r'  and 
near  the  surface,  we  have  r  +  s  practically  equal  to  2?''  and  hence 

v^  =  2g(r  —  s), 

which  is  the  same  as  for  uniform  acceleration  g,  the  initial  speed 
being  zero. 

We  obtained  the  same  result  (page  93)  for  a  body  external  to 
the  earth.  The  equations  of  page  93  hold  good,  therefore,  in  all 
practical  cases,  whether  the  fall  takes  place  above  the  earth  or 
within  the  earth,  neglecting  resistance  of  the  air. 

Amplitude — Epoch — Period — Phase. — The  range  r  =  CA  =  CA' 
on  either  side  of  the  centre  of  acceleration, 
in  harmonic  motion,  is  called  the  amplitude. 

A  complete  oscillation  is  from  A  to  A'  and 
back  to  A.  The  time  of  an  oscillation,  as  we 
have  seen,  is  independent  of  the  amplitude. 
From  A  to  A'  or  A'  to  A  is  a  vibration.  A 
vibration  is  half  an  oscillation.  The  time  of 
a  vibration  is  half  that  of  a  complete  oscillation. 

If  Pi  is  the  initial  position  from  which  the  time  is  counted,  or 
the  position  of  P  at  zero  of  time,  the  time  of  passing  from  A  to  Pi 
is  called  the  epoch.  The  epoch  may  also  be  defined  with  reference 
to  the  auxiliary  circle,  as  the  angle  ACMi  in  radians.  This  is  the 
epoch  in  angular  measure. 

The  epoch  in  angular  measure  is  then  the  angle  described  on  the 
auxiliary  circle  in  the  interval  of  time  defined  as  the  epoch. 

The  epoch  locates  the  position  of  P  at  zero  of  time. 

The  entire  time  which  elapses  from  any  instant  until  the  moving 
point  again  moves  in  the  same  direction  through  the  same  position 
is  called  the  period.  The  time  from  Pi  to  A\  then  back  through 
Pi  to  A,  and  finally  back  from  A  to  P ,  is  a  period.  It  is  evidently 
the  time  of  a  complete  oscillation  from  A  back  to  A. 

That  fraction  of  the  period  which  has  elapsed  since  the  moving 
point  P  last  occupied  A  is  called  the  phase.  Measured  on  the 
circle,  it  is  the  ratio  of  the  angle  ACM  radians  to  2;r  radians. 

Uie  phase  locates  the  position  of  P  at  any  instant. 

It  therefore  varies  with  the  time  or  with  the  position  of  P.  The 
phase  at  zero  of  time,  then,  multiplied  by  2?^  radians  gives  the 
epoch  in  angular  measure,  and  multiplied  by  the  time  of  an  oscilla- 
tion gives  the  epoch  in  time. 

[Application  of  Calculus  to  Harmonic  Motion. — We  may  deduce  tbe 
results  obtained  for  simple  harmonic  motion  (page  104),  as  well  as  others,  from 
the  general  equations  (8)  to  (10),  page  51. 


F 


We  have,  as  before,  a  = —.8  (page  104). 
r 


For  acceleration  away  from   the 

centre  we  have  a  positive,  for  acceleration  towards  the  centre  a  negative  (page 
50). 


106  KINEMATICS   OF   A   POI^TT — TRANSLATION.        [CHAP.  II. 

I.  Acceleration  towards  the  Centre. — In  this  case  we  have 
_  dv  _  cPs  _      a' 

ds 
Multiply  both  sides  by  da  and  then,  since  —  =  v,  we  have 


•vdv  = -sds. 

r 


Integrating,  we  obtain 


^'  =  -^  +  Const.     .    .^  .     .    ,    .    .    (a)        ^ 
lave  then  ^  c    •   ^    ^"^^  /  V  ^  J^ 


When  « =  r  let  «  =  ®i.     We  have  then  ^   z    '   „— -    '*'     •♦  r      f  ^  aj 

•  s  V  '       —     "^ 

Const.  =  ui»  +  ^,  ^' 

and  hence 

«*  =  ^'^  +  7(^' -  «*) (1) 

If  the  initial  velocity  «i  is  zero,  this  becomes 


(2) 


which  is  the  same  as  equation  (1),  page  104,  already  obtained,  for  initial  velocity 

aero  and  range  r. 

ds 
Since  «  =  — ,  we  have  from  (1) 


-»  =  :j7  =  ±  i/  «i  +  -(^  -  «"). 
dt  r  r 


where  we  take  the  (+)  sign  for  motion  away  from  the  centre  and  the  (-)  sign 
for  motion  towards  the  centre  (page  44). 

This  can  be  written  _ 

ds  ,  /a' 

=  =  ±y-,dt (8) 


r  a 

If  we  integrate  this  between  the  limits  of  t  and  <  =  0  when  s  =  r,  we  have 


s  ^  .  /a'  ,     .    _i  r 


Y        '  a'  fa 

s  =  r  cos t  y  -;  ±  Vi  i/  —  sint y  -^ (4) 


*  Let  ^  =  ±t  i/  -„  B=  sin- 


r  a 


«i' 


Then  — - —  =  sin  {A -\- ^  =  sin  A  cos  B  +  cos  A  sin  B. 

r  a 


But  sin  5=  —  -,  and  cos  5=  4/I  —  sin'  B.      Substituting    these 

/— — T—  A 

values  and  reducing,  we  obtain  equation  (4).  ^^  ^at    — •    ]/     ^      i?Cv 


CHAP.  II.] 


SIMPLE  HARMOKIC  MOTION. 


107 


If  motion  is  towards  the  centre,  we  take  the  ( — )  sign ;  if  away  from  the 
centre,  the  (+)  sign. 

If  ©1  is  zero,  or  there  is  no  initial  velocity,  we  have 


«  =  r  cos  t 


^^   ' 


=  \/l' 


If  we  make  u  =  0  in  (1),  we  have  for  the  amplitude 


=  R=  ±  |/r» 


a 


This  reduces  to  ±  r  when  ®i  =  0. 

If  we  integrate  (3)  between  the  limits  of  t  and  <  =  0  when  8  =  B,  that  is,  if 
we  count  the  time  from  the  end  of  the  amplitude  where  Vi  =  0,  instead  of  from 
S  =  r,  we  obtain 


and  hence 


s  =  Bcosty   -;; 


=  4/^ 


-1  s 
^^^     B' 


(7) 


If  in  (5)  .we  make  «  =  —  r  or  in  (7)  make  s  =  —  B,  we  have  in  both  cases 
for  the  time  of  a  vibration,  t  4/  -7-1    and  hence  for  the  time  of  a  complete 


oscillation  T  —  2it 


4/?. 


Therefore  tTie  time  of  oscillation  or  vibration  is  not 


affected  by  the  initial  velocity. 

All  these  equations  (1)  to  (7)  hold  for  motion  either  towards  or  away  from 
the  centre,  provided  the  acceleration  is  towards  the  centre. 

They  also  hold  for  any  path,  straight  or  curved,  provided  r*,  r  and  s  are 
measured  along  the  path  and  a  is  the  rate  of  change  of  speed  at  the  distance  r'. 

II.  Acceleration  Away  from  the  Centre. — In  this  case  we  have  the  accelera- 
tion positive  and  hence 

dv        ,  a' 

dt        '  r 

ds 
Multiplying  by  ds,  we  have,  since  —  =  «, 


Integrating  this,  we  have 


-7-  4-  Const. 


When  «  =  0,  let  c  =  Oi.     Then  Const.  =  Vi^  and 


r 


(8) 


From  (8)  we  have 


ds  .  / 


dt 

('J 


«,'+^A 


^t^       '" 


where  we  take  the  (+)  sign  for  motion  away  and  the  (— )  sign  for  motion 
towards  the  centre.     We  can  put  this  in  the  form 


108  KINEMATICS   OF   A    POINT — TRANSLATION.       [CHAP.  II. 

way  ai 
form 

_v^. 

^^  If  we  integrate  this  between  the  limits  of  t  and  t  =  0  when  «  =  0,  we  have 

logn  («+  j/«''  +  ^,«.»)  =  ±  <|/^  +  logn  vY^,. 
Hence 

(9) 


2  y  a'[  /a'     • 


where  e  is  the  base  of  the  Naperian  system  of  logarithms. 

EXAMPLES. 

g  =  32.16  ft.-per-sec.  per  sec.  or  980.23  cm.-per-sec.  per  sec. 

(1)  If  the  radius  of  the  earth  is  6370900  meters  and  the  accelera- 
tion of  gravity  9.81  meters-per-sec.  per  sec.,  ivhat  should  be  the  valtie 
of  a'r'^  in  eq.  (2),  page  99,  «/  s  and  Si  are  given  in  kilometers  f 

Ans.  398171.88  cubic  kilometers-per-sec.  per  sec. 

(2)  A  body  falls  to  the  earth  from  a  point  1000  miles  above  the 
surface.  Find  its  speed  on  reaching  the  surface,  neglecting  resist- 
ance of  the  air  and  taking  the  eartKs  radius  4000  miles. 

Ans.  ®  =  3.12  miles  per  sec. 


^""^^      (3)  In  the  last  example  find  the  distance  from  the  earth's  surface 
'    ■"  when  the  speed  is  2  miles  per  sec. 
Ans.  535.2  miles. 

(4)  With  uihat  speed  must  a  body  be  projected  vertically  at  the 
earth's  surface  so  that  it  may  never  return  f    (Assume  the  earth  to 
^^^  have  no  atmosphere  and  not  to  be  rotating.) 
....-"^'■v  Ans.  The  speed  is  the  same  as  that  which  a  falling  body  would  have  fall- 

I'^J^       ing  from  an  infinite  distance,  or  «  =  6.95  miles  per  sec. 

,,r^  r^    i     (5)  At  what  point  on  a  line  joining  the  centres  of  the  earth  and 
'  ▼  0  ^'moon  would  the  rate  of  change  of  speed  of  a  body  be  zero  f    (At  the 

*>•    >-   yi^  moon's  surface  a  =  5.5  ft.-per-sec.  per  sec;   radius  of  moon  108O 
i",  ■^}^    miles :  distance  between  centres  of  earth  and  moon  240000  miles.) 
V"^^.  Ans.  Let  a;  =  distance  of  point  from  earth's  centre  and  Xi   from  moon's 

^9'  centre,  and  i2  earth's  radius,  r  moon's  radius.     Then  x-{-Xi  =240000,  and 

,  \.H  ^  "il  ^  ^J^i  .     Hence  x  =  215893  miles. 

^  ar,"         :r 

i  (G)  A  point  whose  motion  is  simple  harmonic  has  velocities  20 
and  25  ft.  per  sec.  at  distances  10  and  8  ft.  from  the  centre  of  ac- 
celeration.    Find  (a)  its  period,  (b)  its  acceleration  at  unit  distance 


from  centre 


.^     ,,.^..,     /->o.'^ 


LaT^^^) 


r 


i 


/-,/> 


-CHAP.  II.]  TRANSLATION — EXAMPLES.  109 

Ans.    We  have  625  =  -,{f  —  64)  and  400  =  ^(rj"  -  100).      Therefore 

T  T 

/a       15  27r         Ait  a's  325 

y  -.  =  -^^^V^^^==~-^  =  -^s^-     a  =  - -r- =  - -5«  =  -  6.25  ft.- 
^      r    ,        b  /^  r  36 

J*  per-sec.  per  sec. 

fi  (7)  7%e  period  of  a  simple  harmonic  motion  is  20  sec.  and  the     ^ 

maximum  velocity  is  10  ft.  per  sec.    Find  the  velocity  at  a  distance 

I  .     of  —  ft.  from  the  mean  position. 

Ans.    — —z  =  20  sec,  therefore  -7  =  -r—-.    Where  «  =  0.  ri  = dt. 

/^  '  ff 

*^  r 

Hence  e"  =  ^(^-;^-  ^1  or  «  =  8  ft.  per  sec.  /  ^     -^,  (>^        ^ / 

'<^S)  A  point  moves  from  rest  towards  a  fixed  point  10  meters  dis-  ^   '■ 

taut,  its  acceleration  being  everywhere  4  times  its  distance  from  a    "^  ^' 
fixed  point.    At  what  distance  will  it  have  a  velocity  of  12  meters 
per  sec.  f 

—-^ Ans.  8  meters. 

(9)  Find  the  mean  speed  of  a  point  executing  a  simple  harmonic 
motion  during  the  time  occupi&l  in  mx)ving  from,  one  to  the  other 
extremity  of  its  range,  its  maximum  speed  being  5  ft.  per  sec. 

Ans.    The  distance  is  2r.      The  time  — -=.      The   mean    speed  ^  ^^ 

Y  r' 

When  8  =  0,  we  have  25  =  —,r^,   or  r  /l/  -,  =  5.    Therefore  mean  speed  is  - — 
r  r    r  It 

ft.  per  sec. 

(10)  If  The  the  period  and  a  the  amplitude  of  a  simple  harmonic 
mixtion,  and  if  v  be  the  velocity  and  s  the  distance  from  the  centre  at 
a  given  instant,  show  that 

(11)  A  point  oscillates  about  a  centre,  its  acceleration  being  pro- 
portional to  its  distance.  Show  that  the  ratio  of  its  maximum  ve- 
locity to  the  square  root  of  the  excess  of  the  square  of  its  maximum 
velocity  over  the  square  of  the  velocity  which  it  has  when  at  a  given 
displacement  from  the  centre  is  equal  to  the  ratio  of  its  maximum 
displacement  to  the  given  displacement. 

(12)  A  point  has  a  simple  harmonic  motion  whose  period  is  4 
trem.  12  sec.    Find  the  time  during  which  its  phase  changes  from  — 

to-  of  a  period. 

Ans.  21  sec. 

[Body  Projected  in  a  Besisting  Medium  —  Acceleration  Proportional  to  the 
Square  of  the  Velocity— Motion  Bectilinear.]— When  a  body  moves  in  a  resisting 


( 


//6^ 


110  KINEMATICS   OF   A   POINT — TRANSLATION.        [CHAP.  II. 

medium  such  as  air  or  water  it  loses  velocity  or  Las  a  minus  acceleration, 
which  is  usually  assumed  to  vary  as  the  square  of  the  velocity. 
We  have  then 

«  =  dr  =  -'^' ^i> 

where  c  is  a  constant  depending  upon  the  shape  and  dimensions  of  the  body 
and  the  density,  or  mass  of  a  unit  of  volume,  of  the  body  and  medium. 

This  constant  is  called  the  coefficient  of  resistance.  For  instance,  for  a 
sphere,  if  d  is  the  diameter  and  A  the  density  of  the  medium  and  8  the  density 
of  the  body,  we  have,  as  is  proved  in  Vol.  Ill,  Kinetics,  page  61, 

_£    J^ 

■^e  can  put  (1)  in  the  form 

^*  ^^  • 

cdt  =  -  -r. 

Let  B  =  z>i  when  t  =  0  and  integrate,  and  we  have 

^B         151/  C  \V  Vi)  ' 

''-T^t (^> 

From  (2)  and  (3)  we  can  find  the  time  for  any  velocity  or  the  reverse.  Since 
«  =  -T-,  we  have  from  (3) 

Integrating  and  making  «  =  0  when  ^  =  0,  we  have 

«=  —  logn(l  +  CTjQ; ■    .    (4) 

c 

or  using  common  logarithms, 

2.302585.      ..   ,        ^^  _ 

«  = Iog(14-CB,0 (5) 

From  (4)  we  have 

t  =  ^- -, (6) 

where  e  —  2.718282  =  base  of  the  Naperian  system  of  logarithms. 

From  (5)  and  (6)  we  can  find  the  distance  for  any  time  or  the  reverse. 
From  (6)  we  have  coxt  =  e"*  —  1,  and  substituting  this  in  (8)  we  have 

•0=^ (7) 

or 

1                 -w 
«=-logn-; (8)' 

or  using  common  logarithms 

2.302585  ,     t), 

*  =  — 7— ^"^7 (^> 

From  (7)  and  (9)  we  can  find  the  velocity  for  any  distance  or  the  reverse. 

From  (8)  we  see  that  when  the  velocity  becomes  zero  the  space  traversed  is 
infinite,  and  from  (2)  the  time  is  infinite.  Although  then  the  velocity  dimin- 
ishes as  the  time  increases,  as  we  see  from  (3),  it  cannot  become  zero  in  any 
finite  distance  or  time. 


r 


CHAP.  II.]  RESISTIKG   MEDIUM.  HI 

[Body  Falling  under  the  Action  of  Gravity  in  a  Kesisting  Medium.]— Let  /be 
the  uniform  acceleration  due  to  any  attractive  force.  In  the  case  of  ffravitv  we 
have*  °         ■' 

(T^^  w^TaLlgJ^  ^  l!^  l-hri  (lepeifcy-op  mass-of-auuit  vulume'Of  tbe-miidlum  and  iS-isHhe 
(teft>;ity  of  the  body.  "The  acceleration / acts  away  from  the  starting-point  and 
tlu'  ntaidution  cv'  acts  towards  the  starting-point.  If  then  we  take  this  point 
as  origin,  we  have 

W^-^-'"' (1) 

where  c  is  the  coeflBcient  of  resistance  and  has  the  same  value  as  in  the  preced- 
ing Article. 

Let  k  be  that  velocity  for  which  the  reta,rdation  is  equal  to/,  so  that/=  cA'^ 

Then  c  =  p,  and  equation  (1)  becomes 

dv  f 


We  can  write  this  in  the  form 


Integrating,  we  have 


:,.        ^*       <^« 


When  <  =  0,  let  t>  =  •»!.    Then  Const.  =  -~.  logn  ^l^i.    Hence 

_  k  {k-^v){k-  «.) 

'-W^''^ik-v){k  +  vS' <*> 

or  using  common  logarithms,  _.. — --  "> 

_  2.3Q2585A;        (*+»)(*  -  ®i) 

' -^sT"^"^  (A -. )(*+.,) !^ 

From  (4)  we  have,  if  e  =  2.718282  =  base  of  Naperian  system  of  logarithms. 


V  = 


P'—e^  +  i 
From  (5)  and  (6)  we  can  find  the  time  for  any  given  velocity  or  the  reverse. 

*  As  we  shall  see  hereafter,  the  mass  of  a  body  multiplied  by  g  gives  the 
weight  of  the  body,  that  is,  the  force  of  gravity.  It  is  also  a  well-known  fact 
that  a  body  immersed  in  a  fluid  has  its  weight  diminished  by  the  weight  of  an 
equal  volume  of  the  medium. 

If  then  V  is  the  volume  of  the  body,  VSg  is  its  weight  in  vacuo  and  V^g 
is  its  loss  of  weight  due  to  the  medium.  Hence  VSg  —  V.^g  is  the  weight 
when  fmmersed.     Since  VS  is  the  mass,  ^  ^ 

rdf=  VSg  -  r^g,    or   f  =  gll-  ~\  .   U^  ^ 


/<:> 
f 


112  KINEMATICS  OF  A  POINT — TEANSLATION.        [CHAP.  II. 

Since  «=  -sr*  we  have  ^  /  -u         1        v 

and  therefore  from  (2)  we  have 

"^    A  ^'  ^'^  iising  common  logarithms, 

^./      U\  /^)  ^  «^  -  ^  2.302585A;' ,      k^  -  ^,^  ,^^ 

K.  k,^  [i,  ^-./,  ^=— 2r-'°^-*^-:^ ^'^ 

'<^<;/^     '^  From'' (7)  we  have 

_  ^ 
««  =  A;»  -  (^2  -  B,s)e     *'   , (9) 

■where  6  =  2.718282  =  base  of  the  Naperian  system  of  logarithms.  < 

From  (8)  and  (9)  we  can  find  the  distance  for  any  velocity  or  the  reverse.        y^S/V" 
I  We  see  from  (9)  that  when  s  is  great,  v  approaches  k,  and  k  is  the  limiting 

value  of  V.  If  the  initial  velocity  is  zero  or  less  than  k,  v  will  continually  ap- 
proach k,  but  can  never  exceed  k.  If  the  initial  velocity  is  greater  than  k,  v 
will  diminish  continually  down  to  k  and  can  never  become  less  than  k.        . 

[Body  Projected  Upwards  under  the  Action  of  ~  Gravity  in  a  Kesisting  Me- 
dium.] 

In  this  case  we  have,  taking  k  as  before,  since /is  negative  and  the  resist- 
ance is  negative, 

''  =  t--f-i^ « 

which  can  be  written  dt  = 35- 1-5— — 5. 

J  «■'-(-«■' 

Integrating  and  determining  the  constant  Ijy  the  condition  that  when  t  =  0, 
«  =  ®j  =  initial  velocity,  we  have 

We  have  also  as  before  -^  =  -t— ,  and  therefore,  from  (1), 
_       k^      vdv 

^^      ^^Integrating  this,  and  making  «  =  0,  when  ®  =  «i ,  we  have  7I  -     /ijL^li/fMf 

Ipiritl: 

2.302585A;2    .       k^-\-'c,^ 

■  *  =  — 2r"   '^  "^M^ ^^^ 

The  time  in  which  the  velocity  becomes  zero  and  the  body  reaches  the  turn- 
ing-point is,  from  (2), 

T=j'---% (5) 

d  the  corresponding  value  of  s  is,  from  (4), 

-       2.302585A;' .       T    ,  ®.n 


CHAP.  II.]  RESISTING   MEDIUM,  113 

At  the  end  of  the  time  T  the  body  begins  to  return  and  falls  from  a  state 
of  rest,  or  Vi  =  0. 

We  have  then  from  the  preceding  Article,  making  Vi  =  0, 

,      _,     2.302585^  ,     k+v  _ 

and 

2.302585A;2,  k^ 

A-.=  -^^log^^--^,  .    ......     (8) 

Let  u  be  the  velocity  with  which  the  body  returns  to  the  starting-point. 
Then  putting  «  =  0  in  (8)  and  v  =  u,  we  have 

or  substituting  for  h  its  value, 

t  =  ~r <») 

Hence 

^-^=F <10) 

We  see  then  that  u  is  less  than  Vi,  or  the  body  returns  to  the  point  of  pro- 
jection with  a  velocity  less  than  the  velocity  of  projection. 

Values  of  —r  and  c. — For  motion  in  a  resisting  medium,  under 

the  action  of  gravity,  we  have/  =  fl'fl  ~  t)>  where ^  is  the  density 

of  the  medium  and  5  that  of  the  body. 

ft 

For  iron  in  water  we  may  take  ^  =       7.2. 

"      "     "    air  "      "        "  ^=5983.28. 

"    mist  or  rain  in  air  "      "        "  -  =   813.82. 

"    lead  in  water  "      "        "  j  =     11-35. 

St 

♦'       "    "  air  "      "        "      2  =9423.61. 

The  coefficient  of  resistance  for  a  sphere  (Vol.  Ill,  Kinetics, 
page  61)  is 

-  L  -  J£. 
*^~Jc'~  16<Jr' 

where  r  is  the  radius  of  the  sphere. 
For  a  cone  we  have 

_       3/Jr'' 

where  r  is  the  radius  of  the  base  and  h  the  height.    If  the  cone 
terminates  in  a  cylinder  of  length  I,  we  have 

_  Sdr" 

^~2d{3l+h){r'  +  h')' 


114  KINEMATICS  OF  A  POINT — TRANSLATION.        [CHAP.  II. 


EXAMPLES.* 

g  =  32.16  ft.-per-sec.  per  sec. 

(1)  A  lead  bullet  1  inch  in  diameter  is  projected  vertically  ivith  a 
velocity  of  2000  ft.  per  sec.  Find  (a)  the  time  of  accent  with  and 
ivithout  resistance  of  the  air ;  (b)  the  distance  to  which  it  ascends 
with  and  without  resistance  of  the  air ;  (c)  the  velocity  and  time  of 
return  with  and  without  resistance  of  the  air. 

Ans.  We  have  in  this  case  %  =  9423.61,  ki  =  67338.1852  and  k  =  259.49 

ft.  per  sec,  /  =  32.1556  ft.-per-sec.  per  sec. 

(a)  The  time  of  ascent  in  vacuo  is  62.19  sec. 

_      .        _,       259.49  ^         ,    2000         ..  ._ 
'^  "■''      ^=  3Tl556*^"  -  '259:49  =  1^-^"  '^- 

(b)  The  height  of  ascent  in  vacuo  is  62189  ft. 

T      .        ,       2.302585  X  67338.1852  /  4000000    \       „-„„  r. 

In  air      h  = ^^  ....„ log     1  4 1  =  8583  ft. 

33.1006  ^    \^^  67338. 1852  y 

(c)  The  velocity  of  return  in  vacuo  is  2000  ft.  per  sec. 

4000000 
In  air    fj.'^  =  ■ 4006{)0(r'  ^^    ^  =  ^57  ft.  per  sec. 


1  + 


67338.1852 


(<?)  The  time  of  return  in  vacuo  is  62.19  sec. 

^      .    ^       ^      2.302585X259.49,       259.49  +  257       „,  ,„ 
In  air  «  -  7=        2x32.1556         ^^^  259.49  -  257  =  ^^'^^  '^*'- 

(2)  A  lead  bidlet  1  inch  in  diameter  is  let  fall  in  the  air.  Find 
the  velocity  at  the  end  of  t  =  1  sec.,  2  sec,  3  sec,  10  sec,  20  sec,  with 
and  without  the  resistance  of  the  air. 

Ans.  We  have  k  =  259.49  ft.  per  sec;  /=  32.1556  ft.-per-sec  per  sec; 

(^        \ 

k\e  *  —  1/ 

e  =  2.718282;  and  from  eq.  (6),  page  111,  making  «i  =  0,  v=  — ^ 

e  ''  + 1 
Let  t  =  1,2,  3,  10  and  20.  and  we  have  d  =  31.98,  63.03,  92.33,  219.3  and 
255.86  ft.  per  sec;    while  in  vacuo  we  would  have®  =  32.16,  64.32,  96.48, 
321.6  and  643.2  ft.  per  sec. 

(3)  In  the  previous  example  what  is  the  greatest  velocity  the  bullet 
can  attain  f 

Ans.  k  =  259.49  ft.  per  sec.  As  we  have  seen,  this  velocity  is  attained 
quite  early,  after  which  the  velocity  is  uniform. 

(4)  An  iron  cannon-ball  1  ft.  in  diameter  is  projected  verticaUi/ 
upwards  in  the  air  with  a  velocity  of  2000  ft.  per  sec.  Find  (a)  the 
time  of  ascent;  (b)  the  distance  to  which  it  ascends  ;  (c)  the  velocity 
with  which  it  returns  ;  (d)  the  time  of  return. 

Ans.  We  have  in  this  case  -,  =  5983.28,  k^  =  513040,  k  =  716.268  ft.  per 
8ec.,f  =  32.1546  ft.-per-sec.  per  sec. 

*  An  examination  of  these  problems  will  give  the  student  an  idea  of  the- 
effect  of  the  air  in  modifying  the  motion  of  a  falling  body  and  enable  him  to 
realize  the  inaccuracy  of  neglecting  it. 


CHAP.  II.]  BESISTINQ  MEDIUM — EXAMPLES.  115 

(a)  The  time  of  ascent  is  T  —  27.32  sec;  in  vacuo,  63.19  sec,  (See  Ex, 
(1).) 

(6)  The  distance  of  ascent  is  h  =  34680  ft.;  in  vacuo,  62189  ft. 

(c)  The  velocity  of  return  is  u  =  6T4.3  ft.  per  sec;  in  vacuo,  2000  ft.  per 
sec 

(d)  The  time  of  return  is  <  —  T  =  88.9  sec;  in  vacuo,  62.19  sec 

(5)  An  iron  cannon-ball  1  ft.  in  diameter  is  let  fall  in  the  air. 
Find  the  velocity  at  the  end  of  t  =  1  sec,  2  sec.,  3  sec,  10  sec,  40  sec, 
60  sec. 

Ans.  We  have  k  =  716.268  ft.  per  sec.,/=  32.1546  ft.-per-sec  per  sec.  and 
Vt  =  0.     Hence  from  eq.  (6),  page  111  : 

For  f  =      1             2             3  10  40  60    sec. 

V  -  32.12        64        95.85  301.5  677.8  709.7  ft.-per-sec. 
In  vacuo  we  would  have 

®  =  32.16        64.32  96.48  321.6  1286.4  1929.6  ft.  per  sec 

(6)  In  the  previous  example,  what  is  the  greatest  velocity  the  can- 
non-ball can  attain  f 

Ans.  k  =  716.268  ft.  per  sec.  And  this  is  attained  in  little  more  than  a 
minute. 

(7)  In  Example  (5)  what  are  the  distances  passed  through  f 

Ans.  From  eq.  (8)  we  have,  making  Ci  =  9  and  taking  the  values  of  « 
already  found. 

For  t=l            2  3              10             40  60    sec. 

8  =  16.03     64.29  144.2  1561  18000  31992  ft. 
In  vacuo  we  would  have 

s  =  16.08     64.32  144,73  1608  25728  57888  ft, 

(8)  A  lead  shot  i  inch  in  diameter  is  let  fall  in  the  air.  Find  the 
greatest  velocity  it  can  attain,  and  the  velocity  and  space  traversed 
in  t  =  1,2,  3,  4,  5,  6,  7  and  8  sec. 

Ans.  We  have  -|-  =  9423.61,/=  32.1556,  k^  =  8417. 

Greatest  velocity  =  A;  =  91.7  ft.  per  sec. 


For  t  =     1 
V  =  30.89 

2 
55.49 

3 

71.76 

4 
81.23 

5     6 

86.35  89.01 

7     8  sec. 
90.35  91.03  ft.  per  sec 

In  vacuo, 
v  =  32.16 

64.32 

96.48 

128.64 

160.8  192.96 

225.12  257.28  ft.  per  sec 

»  =  15.67 

60.27 

123.56 

199 

283.3  370.7 

458.1   545.5  ft. 

In  vacuo, 
16.08 

64.32 

145.72 

257.3 

402   578.8 

787.9  1029.13  ft. 

(9)  What  is  the  greatest  velocity  a  rain-drop  i  inch  in  diameter 
can  acquire,  falling  in  the  air  f 

Ans.  We  have  — p  =  813  82,  /  =  32  12,  A;  =  27  ft.  per  sec. 

(10)  An  iron  cannon-ball  1  ft.  in  diameter  is  let  fall  in  water. 
Find  (a)  the  greatest  velocity  it  can  attain;  (6)  the  final  velocity 
and  the  space  passed  through  in  t  =  1,  2,  3  sec 

Ans.  We  have  -^  =  72,f=  27.7. 

Therefore,  (a)  k  =  23.06  ft.  per  sec 

(6)   For  <  =      1  2  3      sec. 

?)=  19.22        22.68        23.02  ft. per  sec. 
s=1138        32.8  53.94  ft. 

We  see  that  the  maximum  velocity  is  reached  in  about  3  sec. 


116  KINEMATICS   OF   A    POINT — TRANSLATION.        [CHAP.  H, 

In  the  preceding  examples  it  is  assumed  that  the  density  of  the 
medium  is  unchanged,  and  that  the  acceleration  of  gravity  is  con- 
stant. Near  the  earth's  surface  both  assumptions  are  practically 
true.  We  have  also  assumed  that  the  acceleration  varies  as  the 
square  of  the  velocity.  Experiments  would  seem  to  indicate  that 
this  is  not  strictly  accurate. 

The  effect  of  resisting  media  upon  the  motion  of  projectiles  is 
therefore  best  taken  account  of  by  means  of  empirical  formulas 
based  upon  experiment. 

We  have  given  the  preceding  examples  in  order  to  call  attention 
to  the  fact  that  the  influence  of  the  medium,  even  of  the  air,  is  such 
as  to  very  materially  modify  the  results  of  the  formulas  of  page 
93,  which  hold  good  only  in  vacuo  and  are  not  even  approximately 
true  except  for  large  and  heavy  bodies  for  the  first  few  seconds  of 
fall.  The  examples  of  page  95  are  therefore  devoid  of  practical 
value  except  under  such  limitations. 

For  very  ^reat  distances  the  density  of  the  air  and  the  accelera- 
tion of  gravity  are  not  constant,  so  that  our  present  assumptions 
are  then  no  longer  in  accord  with  fact. 


Vi^     /4V/3    ^ 


AV- 


/\Mr^^^^ 


pcj. 


CHAPTER  III. 


TEANSLATION  IN  A  CURVED  PATH 


DIRECTION   OP   ACCELERATION   CONSTANT.      PARABOLIC    MOTION. 
PROJECTILE   IN   A   RESISTING  MEDIUM. 


MOTION     OF 


Curved  Path, — When  a  point  moves  in  a  path  such  that  the  di- 
rection of  the  acceleration  coincides  with  the  direction  of  motion 
and  does  not  change,  the  motion  is  rectilinear,  no  matter  what  the 
law  of  variation  of  the  magnitude  of  the  acceleration  may  be.   Such 
^      motion  we  have  discussed  in  the  preceding  Chapters 
\  If  the  direction  of  the  acceleration,  however,  does  not  coincide 

\     with  that  of  the  motion,  then,  whether  it  is  constant  in  direction 
,    and  magnitude  or  not,  we  have  motion  in  a  curved  path. 

When  a  rigid  body  composed  of  many  points  moves  so  that 
'  every  straight  line  through  any  two  of  its  points  remains  parallel 
to  itself  in  all  positions  of  the  body,  it  has  a  motion  of  trans'lation 
only,  and  we  may  treat  the  body  as  if  it  were  a  point. 

For  motion  in  a  curved  path  the  differential  equations  of  page 

81  apply. 

[  Uniform  Acceleration  Inclined  to  Direction  of  Motion.— If  the  ac- 

\  celeration  is  uniform,  that  is,  constant  in  magnitude  and  direction, 

V  its  component  in  any  given  direction  is  uniform,  and  the  equations 

for  rectilinear  motion,  page  93,  apply  to  the  component  motion  in 

that  direction. 

The  most  common  case  of  curvilinear  motion  under  uniform  ac- 

'      celeration  is  that  of  a  body  projected  with  any  given  velocity  in 

any  given  direction  at  the  surface  of  the  earth,  neglecting  the 

resistance  of  the  air.     In  such  case  the  acceleration  due  to  gravity 

is  practically  uniform  and  equal  to  g  ft.-per-sec.  per  sec. 

;  Let  the  initial  velocity  of  projection  Vi  of  the  point  P  make  the 

I     angle  APB  —  cxi  with  the  hori-  ^^ 

i      zontal. 

\  Let  the  co-ordinates  of  any 

\    point  M  of  the  path  or  trajectory 

\   he  PB  =  x  and  BM  =  y. 

\        Let  the  angle  MPB  =  S.    Let 

\/be  the  uniform  vertical  accel- 

Wation  (in  the  case  of  gravity/ 

V9O: 

-^- — TTwe  make  Op  parallel  and 

equal  to  the  velocity  tJi  at  P,  and 

^       Om  parallel  and  equal  to  the  ve- 

kA       locity  V  at  any  point  M  of  the 

"^  trajectory,  then  ^»i  =  ft  is  the 

integral  acceleration  for  the  time 

t  during  which  the  point  passes  ,   .    ,,      ^   ^ 

from  P  to  M,  and  the  straight  line  pb  is  the  hodograph  (page  52) 

117 


118 


KINEMATICS   OF   A   POIKT— TRANSLATIOIS".       [CHAP.  HI. 


Is 


;^ 


for  motion  from  P  to  the  point  C,  where  the  velocity  is  horizontal, 
and  Ob  is  the  velocity  at  this  point. 

The  velocity  of  p  in  the  hodograph  is  the  acceleration  in  the 
path  (page  52).  Therefore  the  point  p  moves  with  uniform  velocity 
/  from  p  to  b,  while  P  moves  from  P  to  C 

We  see  at  once  that  the  horizontal  component  of  the  velocity  Vi 
is  Ob  =  Vx,  or 

Ob  =  Vx  =  Vi  COB  ai (1) 

The  horizontal  distance  passed  over  in  any  time  t,  while  the 
point  P  moves  from  P  to  if,  is  then  PB  =  x,  or 

PB  =  X  —  Vit  cos  ai (2) 

The  vertical  component  of  the  velocity  Vi  is  bp  =  vi  sin  ai  up- 
wards. But  the  acceleration  /  is  downwards  Hence  the  vertical 
velocity  at  the  end  of  the  time  t  is  bm  =  bp  —  pm  =  Vy,  or 

bm —  Vy=  Vi  sin  ai  — ft (3) 

The  vertical  velocity  at  the  beginning  of  the  time  t  is  Vi  sin  ai . 
The  mean  vertical  velocity  during  the  time  t  is  then 

2Vi  sin  ai  ~  ft  .  1  ^ 
~  =  Vi  sin  ai  —  —ft. 

The  vertical  distance  passed  through  in  the  time  t  is  then 
BM  =  y,  or 

BA-AM=BM~y  =  Vit8in.ai-~ft' (4) 

/*    If  we  combine  (2)  and  (4)  by  eliminating  t,  we  have  for  the  equa- 
tion  of  the  tra^ctory 

i/  =  a;tanari—        /     .^ (5) 

'ivi^  cos^  ai 

This  is  the  equation  of  a  parabola. 
I        The  time  of  reaching  the  highest  point  C  is  the  time  of  describ- 
/  ing  the  vertical  distance  DC.   Call  this  time  Tv .    Since  at  this  point 
\  the  vertical  velocity  is  zero,  we  have,  by  making  %  =  0  in  (3), 

™  _  ■Wi  sinai 
Tv-—j—. 

If  we  substitute  this  for  t  in  (2)  and  (4)  we  obtain  the  co-ordinates 
of  the  vertex  C  of  the  parabola, 

Vi'-'  sin  ai  cos  ori       Ui"  sin  2«j 


PD 
DC 


X,  = 


/ 


2/ 


_       _   t?i*  sin"  ari Xo^ 

~  ^o  ^  2ui'  cos"  «i 


aJo 


The  parameter  of  the  parabola  is  then  - — 


Vi   cos'  ari 


/ 


The  di- 


rectrix is  parallel  to  PD  at  a  distance  above  the  vertex  C  equal  to 

one  half  the  parameter  or  — — -- — - ,  or  at  a  distance  of  -^  above 

P.     That  is,  distance  of  the  directrix  above  P  is  the  height  due  to  the 
velocity  Vi. 

If  we  had  taken  the  origin  at  the  vertex  C  and  let  Xc  and  yc  be 
the  new  co-ordinates,  then  the  horizontal  velocity  at  C  would  be 


■CHAP.  III.] 


TBANSLATIOU — PROJECTILES. 


119 


Vi  COS  a,  and  the  horizontal  distance  passed  over  in  any  time  t  would 
be  Xc  =  Vit  cos  ai.    The  mean  vertical  velocity  would  be  -ft  and 

4i 


the  vertical  distance  yc  —  -ft^ 
obtain 


Combining  and  eliminating  f ,  we 


J      2v^  cos"  ai 
aJc'  = ^ Vc 


Avhich  is  the  equation  of  a  parabola  referred  to  its  diameter  CD  and 

the  tangent  at  the  vertex  C.    The  parameter  is  as  before  — — . 

To  find  the  velocity  at  any  point  of  the  trajectory. — The  magnitude 
bf  the  velocity  at  any  point  M  is  the  resultant  of  the  vertical  and 
liorizontal  velocities,  or,  from  (1)  and  (3), 

i  v'  =  Vx'  +  Vy'  =  Vi^  —  2vift  sin  a,  +f^t' (6) 

The  same  restdt  is  obtained  at  once  from  the  hodograph  from 
the  triangle  0pm. 

Inserting  the. value  of  y  from  (4), 

v^  =  Vi"  -  2fy (7) 

If  the  acceleration  is  due  to  gravity,  we  replace  /  by  gr,  and  have 

— -  = y.    But  we  have  just  seen  that  -—  is  the  distance  of  the 

2g      2g      ^  ''  2g 

directrix  above  P.    Therefore  -^ y  is  the  distance  of  the  direc- 

2g 
if 
trix  above  any  point  M,  and  —  is  the  height  due  to  the  velocity  v. 

Hence,  the  speed  at  any  point  is  the  same  as  that  acquired  by  a  body 
falling  from  the  directrix  to  that  point. 

To  find  the  direction  of  the  velocity  v  at  any  point  M,  the  mag- 
nitude of  which  is  given  by  (6)  and  (7),  let  a  be  the  angle  which  it 
makes  with  the  horizontal.  Then  we  have  directly  from  the  hodo- 
.graph,  since  angle  mOb  =  a, 

■ft; 


vsma  =  Vi  smai 
1}  cos  a  =  Vi  cos  ax. 
Therefore,  from  (2), 

tan  a  =  tan  «i  — 
or 

tan  a  =  tan  ai  — 


ft       _ 


ft' 


=  tan  ai  — 
Vi  cos  ai  X 


fx 


Ui  COS-ai 


(8) 


(9) 


To  find  the  time  of  fiight  in  a  horizontal  direction,  and  the  horizontal 
range. — If  in  (4)  we  make  ^  =  0,  we  have  for  the  time  Th  in  which 
the  body  reaches  the  line  PX,  or  the  time  of  flight  in  a  horizontal 
direction, 

r.  =  ?^^i^J^ (10^ 


J?x, 


f 

Inserting  this  value  of  <  in  (2),  we  have  for  the  horizontal  range 


Bh  = 


2vx  sin  ax  cos  ocx       Vx  sin  2ax 


f 


f 


(11) 


>. 
I 


f 


^ 


120 


V 


iT 


C 


KINEMATICS   OF   A    POINT — TEANSLATION.       [CHAP.  Ill, 


This  is  twice  the  distance  PD. 

The  greatest  value  8in2tri  can  have  is  unity,  and  this  occurs 
when  2«i  =  90°  or  "i  =  45°.  Therefore  the  horizontal  range,  neg- 
lecting resistance  of  the  air,  is  greatest  for  an  angle  of  elevation  of 

45°,  and  is  equal  to  -^ . 

To  find  the  greatest  height  attained,  and  the  corresponding  time. — 
ut  the  vertical  velocity  given  by  (3)  equal  to  zero,  and  we  have 
or  the  time  of  attaining  the  greatest  height 


Tv 


V\  sin  ai 


/ 


(13) 


just  half  the  whole  time  of  flight  as  given  by  (10). 
Insert  the  value  of  the  time  given  by  (12)  in  (4),  and  we  have  for 
le  greatest  height  attained,  CD  =  iJ, 


H^ 


Vi  sin'  ai 


2/ 


(13) 


Equations  (13)  and  -iJft  given  by  (11)  give  the  co-ordinates  of  the 

a 

vertex  C. 

To  find  the  displacement  in  any  given  direction,  and  the  cor- 
sponding  time. — Let  6  be  the  angle  which  any  displacement  FM  = 
makes  with  the  horizontal,  then  we  have  BM  —  y  =  x  tan  ^.  Sub- 
tituting  this  value  of  y  in  (5;,  we  have  for  the  abscissa  of  the 
oint  M 

2ui*  cos  a,  sin  (ai  —  G)  _  f  i''[8in  (2<ari  —  0)  —  sin  6] 


/cose  /cos  6 

a^d  therefore  for  the  displacement  or  range  Pilf  =  R, 

2vi^  cos  ai  sin  (aj  —  S)  _  t^i^Lsin  {2ai  —  6)  —  sin  ( 


R=: 


(14) 


(15) 


/cos'e  /cos^e 

If  in  (15)  we  make  G  =  0,  we  have  the  horizontal  range  Rh  as 
igiven  by  (11). 

If  wie  divide  (14)  by  the  horizontal  component  of  the  velocity, 
Vi  cosai,  we  have 

.1      J.'         ^^-  I.*      2uisin(ai— 6)  ,.„. 

t  =  time  of  flight  = :^— (16) 

'  /cosS 

This  reduces  to  (10)  for  9  =  0. 

To  find  the  angle  of  elevation  which  gives  the  greatest  range  in 
any  given  direction. — The  range  R  given  by  (15)  is  a  maximum  when 

sin  (2a  1  —  6)  is  a  maximum,  or  when2ai  —  9  =  90°  or  a j  =  -  (90°  +  oV 

2  V  / 

The  direction  of  projection  for  the  greatest  range  makes  therefore 

with  the  vertical  an  angle  90  —  ctri  =  -- 1 90  —  9 j,  that  is,  it  bisects  the 

gle  between  the  vertical  and  the  range. 

•To  find  the  elevation  necessary  to  hit  a  given  point. — To  deter- 
mine the  direction  of  the  velocity  Vi  in  order  that  the  path  may 
pass  through  a  given  point  given  by  x  and  y,  we  substitute  for 

— 5 —  the  equivalent  value  1  +  tan'  (n  in  equation  (5),  and  obtain 
cos^  ari  ^ 

t  once 


] 


tan  ai  =  J-  ±  4/  M^    — 
fx     y   \fx) 


1  + 


2u,V 


(17) 


fyv^ 


\    . ~-2_^ ^ —■ -  .\ -7_ 


(l^  ^.  7i    rfi, 4u^ IJ^A"  tlja^i^  iC^  ^  ^^ 


CHAP.  III.]  TRANSLATION — PBOJECTILES. 

Also  from  (14)  we  have 

«-■  =  ^  +  2  sin  -  ^\^    ^,      +  sm  ej, 


2  + 


or,  since  R  cos  B  =  x, 


(18) 


9.1.        JfR  COS' 6  .      .     -\ 
2  +  2  ^^^"T"!^^'" '"'''• 


We  see  from  (17)  that  a,  has  two  values.  If  a/  is  an  angle  such 
that  sin  (2ai'  —  S)  =  sin  i2ai  ~  b),  then  2a/  —  &  =  180"  —  (2(i-i  -  0)  or 
ai'  =  90  —  (ai  —  e)  and  either  a/  or  ni  will  satisfy  equation  (14). 

With  a  given  acceleration  and  initial  velocity  of  projection  of 
given  magnitude,  there  are  therefore  two  directions  of  the  initial 
velocity,  ai  and  90  —  (aj  —  Oi),  and  therefore  two  paths  hy  which  the 
hody  may  attain  the  same  point. 

If  in  (17)  we  put  (?^)'=  1  +  ^^.  we  have 


vi"  =f{y+  Vx'  +  2/" )    and    tan  a,  =  -^r-« 

Smaller  values  of  Vi  make  tan  a,  imaginary.  Larger  values  of 
Vi  give  two  values  for  tan  o-i.  In  the  first  case  the  point  cannot  be 
attained.  In  the  second  case  it  would  be  attained  either  in  the  rise 
or  fall  of  the  projectile. 

To  find  the  envelope  of  all  the  trajectories  corresponding  to 
different  values  of  ai  for  a  given  initial  speed  Vi.  —  Equation  (5) 
gives  the  equation  of  the  trajectory  corresponding  to  the  angle  of 

elevation  ai.    If  we  substitute  1  +  tan"  ai  for        .,     ,  equation  (5) 


becomes 


y  =  x  tan  ai  — 


COS'  ai 
fx\l  +  tan'  ai) 


2vi' 


(19) 


where  x  and  y  are  the  co-ordinates  of  any  point  of  the  path. 

For  another  angle  of  elevation  a/,  and  the  same  initial  speed  Vi, 
we  have 

,      /xi'(l+tan»ai') 
2/,  =  xi  tan  a/  -  ^ 2^ , 

where  Xi  and  yi  are  the  co-ordinates  of  any  point  of  the  new  trajec- 
tory. 


If  we  make  x  =  Xi  and  y  =  yi,  we  have  for  the  point  of  intersec- 
tion of  the  two  trajectories,  by  equating  these  two  equations, 

^(tan  ai  +  tan  a.O  =  1- 

IT  Cro  ^ 


122 


{0 


KINEMATICS   OF   A   POINT — TRANSLATION.      [CHAP.  III. 


If  the  angles  a/  and  ai 
preaches  the  limit 


approach  equality,  this  expression  ap- 


=^tana:i  =1, 


or    tan  ai  = 


(20) 


Equation  (20)  gives  then  the  value  of  tan  a,  when  the  two  tra- 
jectories starting  from  the  same  point  Pwith  the  same  speed  Vi 
have  angles  of  elevation  at  P  whose  difference  is  indefinitely  small. 

Substituting  this  value  of  tan  ai  in  (19)  we  obtain 


(21) 


Equation  (21)  is  then  the  equation  of  a  curve  which  passes 
through  all  the  points  in  which  every  two  trajectories  starting 
from  the  same  point  P  at  angles  of  elevation  whose  difference  is 
indefinitely  small  cut  each  other.  It  is  therefore  the  equation  of 
the  envelope  or  curve  which  touches  all  the  trajectories  or  parab- 
olas described  from  the  same  point  P  with  the  same  initial  speed 

Vu 

Equation  (21)  is  the  equation  of  a  parabola  AC  A',  whose  axis  PC 
is  vertical,  whose  focus  is  the  point  P  of  projection,  and  whose 
vertex  C  is  in  the  common  direction  of  the  trajectories. 

With  the  given  initial  speed  Vi,  the  projectile  can  reach  any 
point  within  this  envelope  by  two  angles  of  elevation  and  two  tra- 
jectories, as  proved  page  121.  It  can  reach  any  point  in  the  en- 
velope by  only  one  elevation  and  path.  It  cannot  reach  with  any 
elevation  and  with  the  given  velocity  Vi  any  point  outside  this 
envelope. 

The  point,  therefore,  where  this  envelope  cuts  the  plane  of  any 
given  range  gives  the  maximum  range  in  that  direction  for  any 
given  Vu 

Thus  the  maximum  range  on  a  horizontal  plane  is  found  from 

Vi' 

(21),  by  making  ?/  =  0,  to  be  -^.    The  same  result  is  given  by  (15) 

when  we  make  ai  =  45°  and  e  =  0. 

Questions  of  maximum  range  may  thus  be  readily  solved  by  the 
equation  for  the  envelope. 

From  (2)  we  have  cos  a  =  —r,  and  from  (4)  sin  ai  =  - — ^ — 

Vit  Vit 

Since  cos"  ai  +  sin'  ai  =  1,  we  have 


x:'  +  {y  +  ^ff]  =vH-' 


(22) 


This  is  the  equation  of  a  circle  whose  radius  is  Vit  and  whose 

centre  is  situated  vertically  below  P  at  a  distance  PD  =  —ff. 

The  circumference  of  this  circle  is  reached  in  the  same  time  by 
a  point  starting  from  P  with  the  velocity  Vi  in  any  direction. 

'Application  of  the  Calculus.]  The  same  results  are  obtained  by  the  applica- 
tion of  the  differential  equations  of  motion,  page  81. 

Thus  in  the  present  case  we  have  for  the  horizontal  component  of  the 
Acceleration,  since /is  vertical, 

»^'' "" 


CHAP.  III.]  TRANSLATIOK— PROJECTILES.  123 

and  for  the  vertical  component 

iPy 

W  =  -^- (») 

Integrating  (a),  since  f or  «  =  0.  ^  =  v,  cos  a„  we  have    '^-/u^   t^o  ,^^:y^e. 

dx             -                                         *'■   ^''    '^   ^^''^> 
^^=at='"''^^''' (1) 

Integrating  again,  since  for  ^  =  0,  a;  =  0,  we  have 

X  =  Vit  cos  OTi (2) 

Integrating  (b),  we  have,  since  when  t  =  0,  :~^  =  Vi  an  a^      ~^  '  *  Z"'^"'  <^  w    '^ 

at  J       1      ^'*-    -^-v  --/T•'-Cw 

^y=^  =  '"^^^^-ft.     .    .  ^    .    '.    ..% 
Integrating  again,  since  ior  t  =  0,  y  =  0,  we  have 

y  =  Vit  sin  ari  -  -^« (4) 

Combining  (2)  and  (4)  by  eliminating  t,  we  have  for  the  equation  of  the 
trajectory 


We  have  also 

or,  from  (1)  and  (2), 

«2  =  •p,"  —  2ftvi  sin  OTi  +/*«« (6) 

Inserting  the  value  of  y  from  (4), 

««  =  tj,2  -  2/y (7) 

If  we  differentiate  (5),  we  have,  for  the  tangent  of  the  angle  which  the 
velocity  at  any  point  makes  with  the  horizontal, 

dy       ^                     fx 
tana  =  -  =  tana,-^^,^^ (8) 

These  are  the  same  equations  as  already  given,  and  from  them  aU  the  others 
are  deduced. 

EXAMPLES. 


g  =  32.16  ft.-per-sec.  per  sec.     Resistance  of  air  neglected. 

(1)  Show  that  for  parabolic  motion  the  hodograph  is  a  straight 
line. 

(2)  Hie  sights  of  a  gun  are  set  so  that  the  ball  may  strike  a  given 
object.  Show  that  when  the  sights  are  directed  to  any  other  object 
in  the  same  vertical  line,  the  ball  will  also  strike  it. 

(3)  Two  bodies  projected  from  the  same  point  in  directions  mak- 
ing angles  /3,  (i'  with  the  vertical  pass  through  the  same  point  in 


124  KINEMATICS   OF   A   POINT — TRANSLATION.       [CHAP.  III. 

the  horizontal  plane  through  the  point  of  projection.    If  t  and  t  are 
the  times  of  flight,  show  that 

sin  {fS  -  /?0  ^  f'  -  f 
sinifJ  +  /f)  ~    r  +  f 

"""^ifWith  what  velocity  must  a  projectile  be  fired  at  an  elevation 
of  30°  so  as  to  strike  an  object  at  the  distance  of  2500  ft.  on  an  ascent 
of  1  in  40  ? 

Ans.  311.5  ft.  per  sec. 

(5)  Find  the  direction  and  magnitude  of  the  velocity  of  projection 
in  order  that  a  projectile  may  reach  its  maximum  height  at  a  point 
whose  horizontal  and  vertical  distances  from  the  starting-point  are 
b  and  h  respectively. 

Ans.  tan  Oi  =  -— ,  v,  =  4/  ^ JT     "^ . 

\JSf^gun  is  fired  horizontally  at  a  height  of  144.72  ft.  above  the 
surface  of  a  lake  and  the  initial  speed  of  the  ball  is  1000  ft.  per  sec. 
Find  (a)  after  what  time,  and  (b)  at  ivhat  horizontal  distance,  the 
ball  strikes  the  lake,  neglecting  resistance  of  the  air. 
Ans.  (a)  3  sec. ;  (6)  3000  ft. 

(7)  In  the  parabola  described  by  a  projectile,  its  speed  at  any 
point  is  that  tvhich  it  would  have  had  had  it  fallen  to  that  point 
from  the  directrix. 

(8)  A  particle  projected  at  a  given  elevation  with  an  initial  speed 
Vi  reaches  the  top  of  a  tower  hft.  high  and  2hft.  from  the  point  of 
projection  in  t  seconds.  Find  (a)  the  initial  speed  of  another  par- 
ticle which,  being  projected  at  the  same  elevation  from  a  point  dis- 
tant 4h  ft.  from  the  tower,  will  also  reach  its  summit,  and  (b)  the 
time  it  will  require. 

Ans.  (a)   -^^^^_.,(,,J^m±K\ 
Vh-\-gt'  Vg 

^-"^j  A  ball  is  projected  with  a  velocity  of  100  ft.  per  sec.  inclined 
75°  to  the  horizon.  Find  (a)  the  range  on  a  horizontal  plane  ;  (b)  the 
range  on  a  plane  inclined  30°  to  the  horizon  ;  (c)  lohat  other  direc- 
tions of  the  initial  velocity  would  give  the  same  ranges. 

Ans.  (a)  155.5  ft.;  {b)  207.3  (V^^ -  1) ;  {e)  15'  and  45°. 

(10)  Show  that  with  a  given  initial  speed  the  greatest  range  on  a 
horizontal  plane  is  just  half  as  great  as  the  greatest  range  doivn  an 
incline  of  30°. 

(11)  Show  that  if  tivo  particles  meet  which  have  been  projected 
with  the  same  initial  speed,  in  tlie  same  vertical  plane,  at  the  same 
instant,  from  tioo  given  points,  the  sum  of  their  elevations  must  b^ 
constant. 

(12)  On  a  small  planet  a  stone  projected  with  a  speed  of  50  ft.  per 
sec.  is  found  to  have  a  maximum  range  on  a  horizontal  plane  of 
400  ft.  Find  the  acceleration  of  falling  bodies  at  the  surface  of  that 
planet. 

Ans.  6  25  ft.-per-sec.  per  sec. 

(13)  Two  stones  thrown  at  the  same  instant  from  points  20  yards 
apart,  with  initial  velocities  inclined  60'  and  30^,  respectively,  to  the 
horizon,  strike  a  flag-pole  at  the  same  point  at  the  same  instant. 


^^1' 


,/^ 


CHAP.  III.]  TRAIfSLATIOlf — PROJECTILES.  125 

Show  that  the  initial  speeds  are  asl:  VB\   and  that  the  distance  of 
th&  pole  from  the  nearer  point  of  projection  is  10  yards. 

^y\\.4c)  At  what  elevation  must  a  body  be  projected  with  a  speed  of 
^^    310.8 /*.  per  sec.  that  it  may  hit  a  balloon  500  ft.  from  the  earth's 
surface  and  at  a  distance  of  1000  ft.  from  the  point  of  projection. 
Aus^^r,  or  80°  43'. 

/  Vfo)  A  body  is  projected  with  an  initial  velocity  of  30  ft.  per  sec. 
inclined  60°  to  the  horizon.    Find  the  velocity  after  20  sec. 

Ans.  617.3  ft.  per  sec.  inclined  148°  36'. 6  to  the  direction  of  the  initial 
velocity. 

(16)  If  from  a  point  A  bodies  are  projected  at  the  same  moment 
and  in  the  same  vertical  plane  at  different  angles  of  elevation,  with 
the  same  initial  speed,  Vi,  the  locus  of  all  the  positions  occupied  at 
the  end  of  a  given  time  t  is  a  circle  whose  radius  is  vit  and  whose 

centre  is  situated  vertically  below  A  at  a  distance  ^gt^. 

J         \,       [Jitt)  A  jet  of  water  rises  vyith  a  velocity  of  20  ft.  per  sec.  at  an 

''^'    angle  of  elevation  of  QQ\    Find  (a)  the  height  due  to  the  velocity; 

(b)  the  greatest  height  of  the  jet;  (c)  the  horizontal  range;  (d)  the 

time  of  reaching  the  horizontal  plane ;  (e)  the  height  corresponding 

to  the  hoHzontal  distance  3  ft. 

Ans.  ^af^MT.;  (b)  5.17  ft.;  (c)  9.24  ft.;  (d)  1.14  sec;  (e)  4.52  ft. 

(18)  A  jet  of  water  discharged  horizontally  at  a  height  above  a 
horizontal  plane  of  lift,  has  a  range  on  the  horizontal  plane  ofH 
ft.    Find  the  velocity  of  projection. 

Ans.  15.92  ft.  per  sec. 

(19)  Prove  that  the  angular  velocity  of  a  projectile  about  the  focus 
of  its  path  varies  inversely  as  its  distance  from  the  focus. 

(20)  Shoiv  that  the  envelope  of  all  the  parabolas  ivhich  correspond 
to  a  given  velocity  of  projection  is  equal  to  the  trajectory  for  which 
the  direction  of  projection  is  horizontal. 

(21)  A  particle  is  projected  over  a  triangle  from  one  end  of  the 
horizontal  base  and,  grazing  the  vertex,  falls  upon  the  other  end  of 
the  base.  If  /3  and  y  are  the  base  angles  and  a  the  angle  of  projec- 
tion, shoiv  that  tan  a  =  tan  /3  +  tan  y. 

(22)  For  the  greatest  range  on  an  inclined  plane  through  the 
point  of  projection  the  direction  of  motion  on  leaving  is  at  right 
angles  tp  that  on  reaching  the  plane. 

1  U***SkA  particle  is  projected  horizontally  with  a  speed  of  32. 16  ft. 

\  vi  per  sec.  from  a  point  128.64 /ee*  from  the  ground.     Find  the  direc- 

tion of  its  motion  when  it  has  fallen  half  way  to  the  ground. 
Ans.  Inclination  to  the  horizontal  =  tan  -  ^  2. 

(24)  The  greatest  range  on  a  honzontal  plane  of  a  projectile 
with  a  given  initial  speed  being  500  meters,  show  that  the  greatest 
range  on  a  plane  inclined  60'  to  the  horizontal  is  2  —  VS  kilometers. 

(25)  A  stone  is  let  fall  in  a  railway-carriage  travelling  at  the 
rate  of  30  miles  per  hour.  Find  its  displacement  relative  to  the 
road  at  the  end  of  0.1  sec. 

Ans.  4.4029  feet,  inclined  2°  5'. 5  to  the  horizon. 


126  KINEMATICS   OF   A    POINT — TRANSLATION.      [CHAP.  III^ 

(26)  The  velocities  of  a  projectile  at  any  two  points  of  its  path 

being  given,  find  the  difference  of  the  altitudes  above  a  norizontal 

plane. 

Fi*  —  Vi^ 
Ans.  : ,  where  V\  and  Fa  are  the  magnitude  of  the  given  velocities. 

(27)  A  given  inclined  plane  passes  through  the  point  of  projection 
of  a  projectile  which  eventually  strikes  the  plane  at  right  angles. 
Find  the  range  of  the  projectile  on  the  inclined  plane,  the  velocity 
of  projection  being  given. 

Ans.  If  6  is  the  inclination  of  the  plane  and  Vi  the  velocity  of  projection, 

■A             •    2«.'        sine 
the  required  range  is ^    ,  „    ■  o  /»  • 

(28)  A  particle  begins  to  slide  from  rest  down  an  inclined  plane 
AB.  At  the  same  instant  another  particle  is  projected  from  A. 
Find  the  condition  that  the  particles  may  meet,  and  ascertain  when 
and  where  this  occurs. 

Ans.  The  second  particle  must  be  projected  at  right  angles  to  the  plane. 
If  S  is  the  inclination  of  the  plane  and  Vi  the  velocity  of  projection,  the  time 

before  meeting  is  <  = ^—r,  and  the  distance  of  the  point  of  meeting  from  A 

°  g  cos  6 

...  ,     2»i  sin  6 

will  be 3-r-. 

g  COS''  0 

(29)  Prove  that  the  components  of  the  velocities  at  the  extremities 
of  any  chord  of  the  path  of  a  projectile,  at  right  angles  to  the  chords 
are  equal. 

(30)  Swift  of  foot  ivas  Hiaicatha  ; 

He  coiild  shoot  an  arroiv  from  him, 

And  run  forward  with  such  jleetness. 

That  the  arrow  fell  behind  him  ! 

Strong  of  arm  was  Hiawatha  ; 

He  could  shoot  ten  aiirows  upward. 

Shoot  them  with  such  strength  and  swiftness, 

That  the  tenth  had  left  the  bow-string 

Ere  the  first  to  earth  had  fallen. 

Supposing  Hiawatha  to  shoot  an  arrow  every  second,  and,  when 
not  shooting  vertically,  to  have  aimed  so  that  the  flight  of  the  arrow 
might  have  the  longest  range,  find  his  speed. 

Ans.   About  99  miles  an  hour. 

(31)  If  any  number  of  bodies  are  projected  from  the  same  point 
in  different  directions  ivith  the  same  initial  speed  Vi,  show  that  the 
foci  of  the  parabolas  they  ivill  describe  will  lie  on  the  surface  of  a 

sphere  whose  radius  is    '  • 
^  2g 

(32)  The  elevation  of  a  projectile  is  that  for  maximum  horizontal 
range.  Find  the  time  of  reaching  a  point  whose  horizontal  and  ver- 
tical distances  from  the  point  of  projection  are  h  and  k  respectively. 

Ans-^^/^i^EI). 

(33)  If  AB  is  the  range  of  a  projectile  on  a  horizontal  plane,  and 
t  the  time  from  A  to  any  point  P  of  the  trajectory,  and  f  the  time 

from  P  to  B,  show  that  the  height  of  P  above  AB  is  ^gtlf. 


CHAP.  III.]  PROJECTILES — RESISTING   MEDIUM.  127 

(34)  A  projectile  is  fired  from  an  elevation  of  1.8  meters  above  a 
horizontal  plane  with  a  horizontal  velocity  of  10  meters  per  sec. 
How  long  before  it  strikes  the  plane  and  what  is  the  range  f(g=  9.81 
meter s-per-sec.  per  sec.) 

Ans.  0.6  sec. ;  6  meters. 

(35)  A  projectile  is  fired  at  an  angle  of  30°  at  a  target  distant  1200 
meters  in  a  horizontal  direction,     (g  =  9.81  meter  s-per-sec.  per  sec.) 

(a)  Find  the  initial  velocity. 

(b)  The  tim^  of  striking. 

(c)  The  highest  point  o/'  the  trajectory. 

(d)  The  velocity  of  striking. 

Ans.  (a)  116.6  meters  per  sec.;  (6)  about  13  seconds;  (c)  173.21  meters;  (d) 
same  as  the  initial  velocity. 

(36)  A  projectile  is  fired  with  an  initial  velocity  of  150  meters  per 
sec.  from  a  point  100  meters  below  a  target  which  is  distant  horizon- 
tally 1525  meters,     (g  —  9.81  meter  s-per-sec.  per  sec.) 

(a)  Find  the  angle  of  elevation. 

(b)  The  velocity  of  striking. 

(c)  The  time  offiight. 

Ans.  (a)  68°  28'  50'  for  bomb,  25°  16'  for  ball;  (6)  143.3  meters  per  sec.; 
(c)  27.715  sec.  for  bomb,  11.34  sec.  for  ball. 

(37)  A  projectile  is  fired  at  an  elevation  of  45°,  and  strikes  a  tar- 
get at  a  horizontal  distance  of  800  meters  aria  70  meters  lower,  (g  = 
9.81  meter  s-per-sec.  per  sec.) 

(a)  Find  the  initial  velocity. 
(6)  The  final  velocity. 

(e)  The  angle  of  striking. 

Ans.  (a)  84.95  meters  per  sec;  (6)  93.6  meters  per  sec;  (c)49°  33'. 

[Motion  of  a  Projectile  in  a  Besisting  Medium.] — In  tbe  preceding  examples 
the  motion  is  assumed  to  be  in  vacuo.  It  should  be  borne  in  mind  that  the  re- 
sistance of  the  air  completely  changes  the  results  of  the  theoretic  formulas. 

The  motion  of  a  projectile,  taking  into  account  the  resistance  of  tbe  air,  i» 
best  given  by  empirical  formulas  based  upon  experiment. 

If,  however,  we  assume  that  the  magnitude  of  the  acceleration  decreases 
directly  with  the  square  of  the  velocity,  we  may  deduce  by  means  of  our  gen- 
eral differential  equations,  page  81,  Chap.  VIII,  the  equation  of  the  trajectory 
for  very  small  angles  of  elevation.  The  same  method  which  we  shall  use 
holds  good  for  any  other  assumption  as  to  tbe  law  of  acceleration. 

The  assumption  is  not  strictly  accurate,  but  will  serve  to  illustrate  the 
method  of  deduction. 

Let  /  be  the  constant  acceleration  due  to  gravity,  the  value  of  which  is 
given  page  111,  and  let  c  be  the  coefficient  of  resistance,  so  that  c  has  the 
value  given  page  113. 

Thus,  since  the  velocity  at  any  point  is  ®  =  — ,  we  have  the  retardation 

— -1    by  assumption.     We  have  then  from  equation   (6),   page  81, 
dt)      ' 

since  cos  a=  — ,  for  the  horizontal  component  of  the  tangential  acceleration 
ds 

d?x  _      /^*V-?-  n\ 

W  -~\dt)di ^^ 

and,  since  sin  a  =  —-,  for  the  vertical  component  of  the  tangential  accelera- 


ds 


d^ 
dp 


=  -/-<^)l <^> 


128  KINEMATICS   OF   A   POINT — ^TRANSLATION.      [CHAP.  III. 


Dividing  (1)  by  — ,  we  have 


at        ax 


Integrating  this,  we  have 

ncrn 

dt 


logn  ^  =  -  cs-\-C, (3) 


where  C  is  a  constant  of  integration.     Let  Vi  cos  nrj  be  the  component  of  the 
initial  velocity  Vi,  parallel  to  x,  ai  being  the  angle  of  elevation  at  the  point  of 

dx 
projection;  then  when  «  =  0,  —  =  Vi  cos  ai,  and  0  =  logn  Vi  cos  a.       There- 
fore, from  ^3), 

dx        _c. 

^  =  e       ®i  cos  tti, (4) 

where  e  is  the  base  of  the  Naperian  system  of  logarithms,  2,718282.     Hence 

dy       dy    dx        -cs  dy 

-jT  =  ^-  Tr  =  ^     «i  cos  «!  -^ (5) 

dt       dx    dt  dx 

If  we  multiply  (2)  by  dx,  and  (1)  by  dy,  and  subtract,  we  have 

^y^--f-^y  =  _jax (6) 

Inserting  the  value  of  dt^  from  (4),  we  have 

d?ydx  -  d^xdy  _  _      fe^      ^^  ^         ^. 

dX^  Vi^  COS'^  OTi 

dv 
The  first  member  of  (7)  is  equal  to  d-^.     We  have  also 

/.    .  dy^\\       ds  ^  da 

dx^  +  dy^  =  ds\     or     {l+£)    =-^.    or    ^^  = 

Substituting  these  values  in  (7),  we  have 

i^  +  d^J  ^^--^.^cos^a,'^ ^^> 

If  we  assume  the  angle  of  projection  a,  as  very  small,  so  that  the  trajectory 
is  very  flat,  we  have  approximately  in  such  case 

ds  =  dx,     and    «  =  a;,     and     II  +  ;^)  =  1* 


Therefore  (8)  becomes 


af=-^^dx (9) 

dx  Vi^  cos^  ai 


Integrating  (9),  since  when  x  =  0,  -p-  =  tan  ai ,  we  have 


^=tana.-^-/-— (e^-1) (10) 

dx  2c»i*  cos''  r'  ^  '  "^    ' 


Integrating  (10),  we  have,  since  for  a;  =  0,  j'  =  0, 

«  =  a!tanai-f  ir— 5--^^-^ .  „    /   ,      (e^-  iV    .     .(11) 

*  '   2cvi^ cos**  aj       4c'»i* cos*  ai\  J  ^    ' 


CHAP.  III.]  PROJECTILES — RESISTING   MEDIUM.  129 

Equation  (11)  is  the  approximate  equation  of  the  path. 
If  we  expand  the  last  term  in  a  series,. we  have 

y  =  «tan  a, r- ^ —  (^9\ 

If  the  terms  containing  c  are  omitted,  and /=  fir,  equation  (12)  is  that  of  a 
parabola,  which  is  the  path  of  the  projectile  in  vacuo. 

The  ordinate  of  the  actual  curve  is  therefore  less  at  any  distance  x  than  tha'c 
of  the  parabola  for  the  same  distance. 

From  equation  (4)  we  have 

—r-  =  e~'^  «,  cos  «! ; 

and  integrating,  since  for  a;  =  0,  <  =  0,  we  have 

cvi  cos  ait  =  e""  -  1, (13) 

which  gives  the  time  in  terms  of  the  abscissa. 
From  (13)  we  have 

X  =  — logn(ciJi  cosait-{-l), 

....    (14) 


or,  in  common  logarithms, 

2.302585 


X  = 


c 


log(cPiCosa,f-f  1), 


which  gives  the  abscissa  x  in  terms  of  the  time. 

"The  general  problem  of  the  path  of  a  projectile  in  a  uniform  resisting 
medium,  where  the  resistance  varies  as  the  square  of  the  velocity,  was  pro- 
posed by  Keill  as  a  trial  of  skill  to  John  Bernoulli,  by  whom  the  challenge  was 
received  Feb.  1718.  Keill,  trusting  to  the  complexity  of  the  analysis,  which 
had  probably  deterred  Newton  from  attempting  any  regular  solution  of  the 
problem  in  the  second  book  of  the  Principia,  was  in  hopes  that  the  exertions  of 
Bernoulli  would  prove  unsuccessful.  Bernoulli,  however,  having  expeditiously 
effected  a  solution,  not  only  of  Keill's  problem,  but  likewise  of  the  more  general 
one  where  the  resistance  varies  as  the  nth  power  of  the  velocity,  expressed  a 
determination  not  to  publish  his  investigation  until  he  had  received  intimation 
that  his  antagonist  had  himself  been  able  to  solve  his  own  problem.  He  gave 
Keill  till  the  following  September  to  exercise  his  talents,  declaring  that  if  he 
received  by  that  time  no  satisfactory  communication,  he  should  feel  himself 
entitled  to  question  the  ability  of  his  adversary.  At  the  request  of  a  common 
friend,  Bernoulli  consented  to  extend  the  interval  to  the  first  of  November. 
It  turned  out,  however,  that  Keill  was  unable  to  obtain  a  solution.  At  length 
Nicholas  Bernoulli,  Professor  of  Mathematics  at  Padua,  communicated  to  John 
Bernoulli  a  solution  of  Keill's  problem,  which  the  author  afterwards  extended 
to  the  more  general  one.  Finally,  on  the  17th  of  November,  information  was 
received  by  John  Bernoulli  from  Brook  Taylor,  to  the  effect  that  he  had  ob- 
tain-d  a  solution.  John  Bernoulli  published  his  own  analysis,  together  with 
that  of  his  nephew  Nicholas,  in  the  Acta  Erudit.  Lips.  1719  mai.,  p.  216." 
(Walton,  Problems  in  Theoretical  Mechanics.) 


CHAPTER  rv. 

CURVILINEAR  MOTION  OF  TRANSLATION— CENTRAL 
ACCELERATION.    HARMONIC  AND  PLANETARY  MOTION. 

Central  Acceleration. — If  the  acceleration  of  a  moving  point  i» 
always  directed  towards  a  fixed  point  or  centre  of  acceleration,  the 
acceleration  is  said  to  be  central. 

The  velocity  of  the  moving  point  at  any  instant  is  the  resultant 
of  the  velocity  at  the  preceding  instant  and  of  the  integral  accelera- 
tion during  the  intervening  time. 

But  if  the  acceleration  is  always  directed  towards  the  centre,  or 
fixed  point,  its  moment  with  reference  to  that  point  is  zero.  Since 
the  moment  of  the  resultant  of  any  two  components  about  any 
point  is  equal  to  the  sum  of  the  moments  of  the  components,  and 
since  in  this  case  the  moment  of  one  of  the  components,  viz.,  the 
acceleration,  is  zero,  it  follows  that  the  moment  of  the  velocity  about 
the  centre,  in  the  case  of  central  acceleration,  is  constant. 

Conversely,  if  the  moment  of  the  velocity  of  a  moving  point 
about  any  fixed  point  is  constant,  the  acceleration  must  always  be 
directed  towards  that  point. 

If  r  is  the  distance  of  the  moving  point  from  the  fixed  point,  p 
the  lever-arm,  and  go  is  the  angular  speed  at  any  instant,  we  have 
for  the  moment  of  the  velocity  pv  —  r'ao  =  c,  equal  to  twice  the 
areal  velocity  of  the  radius  vector  (page  76). 

Therefore,  in  all  cases  of  central  acceleration -r'ca  is  constant,  or 

a 

the  area  described  by  the  radius  vector  in  a  unit  of  time  is  constant. 

c 
It  follows  also  that  in  all  cases  of  central  acceleration  ca  =  - ,  or 

the  angular  speed  is  inversely  as  the  square  of  the  radius  vector. 

Cases  of  Central  Acceleration. — The  two  most  important  cases 
of  central  acceleration  are  those  of  harmonic  motion,  where  the 
central  acceleration  is  directly  proportional  to  the  distance  from  the 
centre,  and  planetary  motion,  where  it  is  inversely  as  the  square  of 
the  distance. 

When  the  velocity  is  in  the  same  straight  line  as  the  central 
acceleration  we  have  in  both  these  cases  rectilinear  motion.  The 
first  is  simple  rectilinear  harmonic  motion,  the  second  is  rectilinear 
planetary  motion  or  that  of  a  body  at  great  distances  from  the 
earth.     Both  these  cases  have  been  considered  in  Chaps.  I  and  II. 

When  the  velocity  is  not  in  the  same  straight  line  with  the 
central  acceleration  we  have  compound  harmonic  motion  and 
planetary  motion  in  general.  The  first  is  of  great  importance  in 
the  study  of  sound,  light,  heat,  etc.,  as  well  as  in  ordinary  kinetics. 
The  second  is  the  motion  of  planets  about  the  sun  and  of  satellites 
about  their  primaries. 

130 


CHAP.  IV.]   TRANSLATION — COMPOUND  HARMONIC  MOTION. 


131 


Cases  of  Harmonic  Motion. — We  have  defined  simple  harmonic 
motion,  page  103,  as  the  motion  of  a  point  moving  in  any  path  in 
such  a  manner  that  the  tangential  component  of  the  acceleration  a 
is  directly  proportional  to  the  distance,  measured  along  the  path» 
from  a  fixed  point  in  the  path. 

Such  motion  may  be  rectilinear  or  curvilinear.  In  the  first  case 
it  is  simple  rectilinear,  in  the  second  simple  curvilinear. 

If  the  whole  acceleration  itself,  or/,  is  central,  that  is,  always 
directed  towards  a  fixed  point  not  in  the  path,  and  is  always  propor- 
tional to  the  distance  from  this  fixed  point,  the  motion  is  central 
harmonic,  or  compound  harmonic,  so  called,  because  it  is  the  result- 
ant of  two  simple  rectilinear  motions,  as  will  be  proved  in  the  next 
article. 

Simple  rectilinear  harmonic  motion  is  also  central,  because  the 
fixed  point  is  in  the  path. 

Any  Central  Harmonic  Motion  may  be  Resolved  into  Two  Simple 
Rectilinear  Harmonic  Motions  at  Right  Angles. — Let  C  be  the 
centre  of  acceleration,  and  P  the  position  of  the  moving  point  at 
any  instant.  Let  the  velocity  u  of  P  make 
an  angle  a  with  the  axis  of  X,  and  let  the 
motion  of  P  be  harmonic  so  that  the  ac- 

a' 
celeration  of  P  is  —rV,  where  a'  is  the  ac- 

r 
celeration  at  a  known  distance  r',  and  r 
is  the  distance  CP. 

The  velocity  v  may  be  resolved  into 
V  cos  a  and  v  sin  a  in  the  directions  GX 

a' 

and  CY,  and  the  acceleration  may  be  resolved  into  —rV  cos  PC  A  or 

r 

— rCA,  and  -j-r  cos  PCB  or  -j^CB,  in  the  same  directions. 

The  component  accelerations  are  therefore  directly  as  the  dis- 
tances CA  and  CB,  and  the  component  velocities  are  in  the  direc- 
tions of  CA  and  CB.  The  central  harmonic  motion  of  P,  whatever 
the  direction  of  the  velocity  v,  is  therefore  the  resultant  of  two 
simple  harmonic  motions  in  the  lines  CA  and  CB  at  right  angles. 

If  then  any  central  harmonic  motion  is  resolved  into  two  com- 
ponents at  right  angles,  the  component  motions  are  rectilinear  har- 
monic. 

Conversely,  the  resultant  of  two  rectilinear  harmonic  motions  at 
right  angles  is  a  central  harmonic  motion.  Central  harmonic  mo- 
tion is  therefore  called  compound  harmonic  motion. 

Composition  of  the  Simple  Rectilinear  Harmonic  Motions  in  Dif- 
ferent Lines. — ^Let  the  point  M  move  in  a  circle  AMA'  of  radius  r  = 

CA  =  CM  with  a  constant  angular 
velocity  go.  Then  the  motion  of  the 
projection  P  in  the  line  AA  is 
simple  rectilinear  harmonic  (page 
103). 

Let  the  point  M,  move  in  the 
circle  CBMi  of  radius  ri  =  CB  ~ 
Ja  CMi ,  with  constant  angular  veloci- 
ty oai.  Then  the  motion  of  the  pro- 
jection Pi  in  the  line  CB  is  simple 
rectilinear  harmonic.  Let  the  angle 
BCA  between  the  planes  of  the 
circles  be  a. 


Via.  1. 


133  KINEMATICS   OF   A   POINT — TKANSLATION.       [CHAP.  IV. 

Let  the  time  count  from  the  instant  when  M,  is  at  B,  so  that  the 
epoch  of  Pi  is  zero  (page  105).  At  this  instant  let  the  epoch  of  P  be 
€.  Then  e  is  the  difference  of  epoch,  or,  in  angular  measure,  the 
angle  of  ilf  above  or  below  A  at  the  beginning  of  the  time.  In  any- 
time t.  Ml  will  have  moved  from  B  through  the  angle  wit  measured 
from  CB,  and  M  through  the  angle  cot  ±  e  measured  from  CA. 

By  the  preceding  Article  we  can  resolve  the  harmonic  motion  of 
Pi  into  a  simple  rectilinear  harmonic  motion  at  right  angles  to  CA, 
-and  another  along  CA. 

The  displacement  of  Pi  from  C  for  any  time  t  is  Vi  cos  (oo,t),  and 
this  displacement  may  be  resolved  into  vi  cos  a  cos  (ood)  along  CA, 
and  Vi  sin  a  cos  (osit)  perpendicular  to  CA.  The  displacement  of  P 
from  C  in  the  same  time  t  is  r  cos  (cot  ±  e). 

If  a  point  undergoes  these  displacements  simultaneously,  its  re- 
sultant displacement  along  CA  wUl  be 

X  =  r  cos  {oot  ±  e)  +  Vi  cos  a  COS  (ttJif),    ....     (1) 
and  perpendicular  to  CA 

y  =  Vi  sin  acos  (ooit) (2) 

The  equation  of  the  curve  in  which  the  point  moves,  referred  to 
rectangular  co-ordinates  with  C  for  the  origin,  will  then  be  obtained 
by  combining  (1)  and  (2)  so  as  to  eliminate  t.  Such  combination 
(page  131)  gives  always  a  central  or  compound  harmonic  motion 
about  C,  the  radius  vector  from  C  passing  over  equal  areas  in  equal 
times  (page  130). 

Equations  (1)  and  (2)  enable  us  then  to  find  the  curve  resulting 
from  the  combination  of  any  two  simple  rectilinear  harmonic  mo- 
tions inclined  at  any  angle  a. 

If  the  component  motions  are  at  right  angles,  a  =  90°.  If  the 
amplitudes  are  equal,  r  =  ri.  If  the  periods  are  equal,  go  =  an,  the 
difference  of  epoch  is  constant,  and,  since  the  epoch  equals  the  pro- 
duct of  the  phase  at  zero  of  time  by  Stt  radians  (page  105),  when  the 
periods  are  equal  the  difference  of  phase  is  constant.  When,  then, 
the  periods  are  equal  and  e  =  0,  or  the  epochs  are  equal,  the  phases 
are  also  equal  at  any  instant.  (For  definitions  of  amplitude,  period, 
epoch  and  phase,  see  page  105.) 

Two  Component  Simple  Rectilinear  Harmonic  Motions  in  Differ- 
ent Lines  with  the  Same  Period. — In  this  case  go  —  go,  and  e  is  con- 
stant, or  the  difference  of  epochs  is  constant  and  difference  of 
phase  at  any  instant  is  constant. 

We  have  then,  from  (1)  and  (2), 

x  =  r  cos  (cot  +  e)  +  ri  cos  a  COS  (aot),  y  =  r  sin  a  cos  {cot). 
Combining  these  two  equations  by  eliminating  oot,  we  have 

(ri"  sin"  a)x'  -  2ri  sin  a{r  cos  e  +  r,  cos  a)xy  + 

(r"  +  2rri  cos  <?  cos  a+  ri"  cos"  «)!/"  =  rVi"  sin"  a  sin"  e.     (3) 

This  is  the  equation  of  an  ellipse  referred  to  its  centre  and  rect- 
angular axes. 

Hence  if  a  point  has  two  component  simple  rectilinear  harmonic 
motions  in  any  directions,  of  any  amplitudes,  and  any  difference  of 
epoch,  if  the  periods  of  the  two  components  are  the  same,  the  result- 
ant motion  of  the  point  will  be  central  harmonic  in  an  ellipse,  the 
centre  of  acceleration  at  the  centrt:.  of  the  ellipse.  The  areal  velocity 
of  the  radius  vector  about  the  centre  is  constant  (page  130). 

Such  motion  is  called  elliptic  harmonic  motion.    Elliptic  har- 


CHAP.  IV.J    TRANSLATIOSr — COMPOUND  HARMONIC  MOTION. 


133 


monic  motion,  then,  is  compound  harmonic  motion  when  the 
periods  of  the  components  are  the  same. 

Equation  (3)  gives  all  cases  of  compound  harmonic  motion  for 
equal  periods  of  the  components. 

It  will  be  instructive  to  derive  from  it  special  cases. 

(a)  Two  Component  Simple  Rectilinear  Motions  in  Different  Lines 
with  the  Same  Period  and  Phase. — In  this  case  we  make  in  (3)  e  =  0, 
and  therefore  the  phases  are  equal,  and  we  have  at  once 

r  +  ri  cos  a 

X  — . y. 

Ti  sm  a 

This  is  the  equation  of  a  straight  line  passing  through  the  centre 
C.     The  resultant  motion  is  therefore 
central  harmonic  in  a  straight  line,  or 
simple  rectilinear  hai'monic. 

If  CA  and  CB  are  the  amplitudes  r 
and  ri  inclined  at  the  angle  cr,  the  result- 
ant motion  has  the  amplitude  CR,  in        ^  y  a 
direction  and  magnitude  the  diagonal  of                       Fiq.  2. 
the  parallelogram  whose  adjacent  sides  are  r  and  vi,  inclined  at 
the  angle  a. 

Conversely,  a  simple  rectilinear  harmonic  motion  whose  ampli- 
tude is  CR  may  be  resolved,  by  completing  the  parallelogram,  into 
two  others  in  any  two  directions,  of  the  same  period,  epoch  and 
phase. 

Ti 

If  a  =  90°,   we  have  y  =  —x.    Therefore  the  projection  of  a 

simple  rectilinear  harmonic  motion  on  any  straight  line  is  also 
a  simple  rectilinear  harmonic  motion  of  the  same  period,  epoch 
and  phase. 

If  the  component  motions  are  more  than  two,  they  may  be  com- 
pounded two  and  two,  and  therefore  any  number  of  component 
simple  rectilinear  harmonic  motions  in  any  directions,  of  the  same 
period,  epoch  and  phase,  give  a  single  resultant  rectilinear  har- 
monic motion  of  determinate  direction  and  amplitude,  which  may 
be  resolved  into  two  components  in  any  two  directions,  of  the  same 
period,  epoch  and  phase. 

(b)  Two  Component  Simple  Rectilinear  Motions  in  the  Same  Line 
with  the  Same  Period  and  Different  Epochs  and  Phases. — In  this  case 

R    we  make  in  (3)  a  =  0,  and  obtain 
at  once 

(r*  +  2rri  cos  e  +  r^)y^  =  0. 

But  since  for  a  =  0,  y  =  0,  (see 
Fig.  1,)  we  have 

r*  +  2rri  cos  e  +  ri"  =  constant. 

In  Fig.  3  the  points  P  and  Pi 
move  in  the  line  AA'  with  sim- 
ple harmonic  motion  and  the 
diagonal  CR  =  i^r^^  2rr,  cos  e  +  r.',  where  e  is  the  constant  differ- 
ence of  epoch  and  phase.  _  . 

Since  e  is  constant  and  CR  is  constant,  its  mclmation  to  CM  or 
CMi  is  constant.  At  any  instant  the  resultant  displacement  is 
CPi  +  CP  =  CS,  and  the  motion  of  S  is  therefore  the  resultant 
motion  and  is  simple  rectilinear  harmonic,  with  the  amplitude  CR, 
the  diagonal  of  the  parallelogram  on  r  and  r..    The  epoch  and 


134  KINEMATICS  OF  A  POINT — TRANSLATION.       [OHAP.  IV. 

phase  are  intermediate  between  the  epochs  and  phases  of  the  com- 
ponents. 

If  the  epochs  and  phases  are  the  same,  e  =  0  and  the  ampHtude  of 
the  resultant  motion  is  r  +  ri ,  or  the  sum  of  those  of  the  compo- 
nents. If  the  difference  of  epoch  or  phase  is  e  =  tt  radians,  the  am- 
plitude is  r  —  ri  or  the  difference  of  those  of  the  components. 

By  taking  CMi  and  CM  of  proper  lengths  we  can  make  MCP  and 
MiCM  what  we  please  without  changing  CR.  Therefore  any  simple 
rectilinear  harmonic  motion  m.ay  be  resolved  into  tivo  others  in  the 
same  line,  with  any  required  difference  of  phase  and  one  of  them 
having  any  desired  epoch,  the  periods  being  the  same. 

Three  or  more  component  simple  rectilinear  harmonic  motions  in 
the  same  line  and  with  the  same  period  may  be  compounded  two 
and  two,  and  the  resultant  will  be  rectilinear  harmonic  with  the 
same  period. 

If  the  periods  are  different,  the  angle  MiCM=  e  will  vary  and 
CR  will  vary.  When  e  =  0,  CR  will  have  its  maximum  value  r-\-ri. 
"When  the  difference  of  epoch  e  is  tt  radians,  CR  has  its  minimum 
value  J —  ?'i.  The  angular  velocity  of  CR  is  also  variable.  The  di- 
rection of  CR  will  oscillate  back  and  forth  about  CM,  the  maximum 

1  r\ 
inclination  being  sin       — .    The  resultant  motion  is  therefore  not 

simple  rectilinear  harmonic,  but  a  more  complex  motion.  It  is,  as 
it  were,  simple  harmonic  with  periodically  increasing  and  decreas- 
ing amplitude,  and  periodical  acceleration  and  retardation  of  phase, 
or  epoch. 

(c)  Two  Component  Simple  Rectilineax  Harmonic  Motions  at  Right 
Angles  with  the  Same  Period  and  Different  Phases  or  Epochs. — The 
general  equation  for  this  case  is  given  by  (3).  If  the  directions 
are  at  right  angles,  we  have  a  =  90°.  Suppose  in  addition  the  am- 
plitudes equal,  so  that  r  =  ri,  and  the  difference  of  epoch  e  =  90°. 
We  have  then,  from  (3), 

Since  the  motion  is  central  harmonic,  according  to  page  130  the 
areal  velocity  of  the  radius  vector  is  constant ;  and  since  the  radius 
is  constant,  the  speed  in  the  circle  is  constant.  We  have  already 
seen,  page  103,  that  the  projection  of  the  motion  of  a  point  moving 
with  uniform  speed  in  a  circle,  upon  a  diameter,  gives  rectilinear 
harmonic  motion.  The  projection  upon  two  diameters  at  right 
angles  gives  then  two  component  rectilinear  harmonic  motions  of 
the  same  period,  with  a  difference  of  epoch  of  90°,  or  of  phase  of 
i,  since,  when  one  component  has  its  greatest  displacement,  the 
other  is  at  the  centre  with  displacement  zero. 

It  follows  also  that  two  component  simple  rectilinear  harmonic 
motions  at  right  angles,  with  the  same  period  and  equal  amplitudes, 
differing  in  epoch  by  90°  or  in  phase  by  one  quarter  of  a  period,  will 
give,  as  a  resultant,  uniform  motion  in  a  circle  whose  radius  is  the 
common  amplitude  of  the  components. 

If  the  amplitudes  are  not  equal,  but  oc  and  e  still  90°,  and  periods 
the  same,  we  have,  from  (3), 

ri'ar'  +  r^y^  =  rVi", 

which  is  the  equation  of  an  ellipse  referred  to  its  centre  and  axes. 
The  resultant  motion  is  therefore  central  harmonic  in  an  ellipse, 
whose  semi-diameters  are  r  and  r\,  the  centre  at  the  centre  of  the 
ellipse. 


CHAP.  IV.]   TKANSLATION — COMPOUND  HARMONIC  MOTION.  135 

The  same  result  is  evidently  obtained  by  projecting  the  circle  in 
the  preceding  case  upon  a  plane,  so  as  to  obtain  the  required  ampli- 
tude ri,  r  remaining  unchanged. 

id)  Three  or  More  Component  Simple  Rectilinear  Haxmonic  Mo- 
tions in  Different  Lines  with  the  Same  Period  but  Different  Phases  or 
Epochs. — We  have  seen  from  equation  (3)  that  the  resultant  of  two 
simple  rectilinear  component  harmonic  motions  in  any  two  direc- 
tions, of  the  same  period  and  different  epoch  or  phase,  is  elliptic 
harmonic  motion. 

We  have  also  seen  from  (a)  that  any  simple  rectilinear  harmonic 
motion  may  be  resolved  into  two  others  of  the  same  period  and 
phase  or  epoch  in  any  two  given  directions.  Any  niimber  of  given 
simple  rectilinear  harmonic  motions,  then,  of  the  same  period  and 
different  phases  or  epochs  may  each  be  resolved  into  its  own  pair 
in  any  two  given  directions.  These  pairs  constitute  a  number  of 
simple  rectilinear  harmonic  motions  in  two  given  lines,  all  of  the 
same  period  and  different  phases  or  epochs. 

According  to  (6),  all  in  one  line  may  be  compounded  into  one  re- 
sultant, and  all  in  the  other  line  into  another  resultant,  these  two 
resultants  having  the  same  period  and  different  phases  or  epochs. 
The  resultant  of  these  two  is,  according  to  equation  (3),  elliptic  har- 
monic motion. 

Hence  the  resultant  of  any  number  of  component  simple  rectili- 
near harmonic  motions  of  the  same  period,  whatever  their  ampli- 
tudes, directions,  phases  or  epochs,  is  elliptic  harmonic  motion, 
the  centre  of  the  ellipse  being  then  centre  of  acceleration,  and  the 
radius  vector  describing  equal  areas  in  equal  times. 

In  special  cases  this  becomes,  as  we  have  seen,  uniform  circiilar 
motion  or  simple  rectilinear  harmonic  motion. 

Since  the  above  holds  whatever  the  inclination  of  the  two  result- 
ants, elliptic  harmonic  motion  may  be  considered  as  the  resultant 
of  two  component  simple  harmonic  motions  of  the  same  period 
and  different  epochs  or  phases  at  right  angles. 

Graphic  Representation. — We  may  exhibit  graphically  simple 
or  compound  rectilinear  harmonic  motion  by  a  curve  in  which 
the  abscissas  represent  intervals  of  time,  and  the  ordinates 
the  corresponding  distance  of  the  moving  point  from  its  mean 
position.  . 

In  the  case  of  a  single  harmonic  motion  we  have  (page  16£) 
jc  =  r  cos  {oot  ±  e).     If  the  distance  x  is  to  be  zero  when  *  =  0,  we 

must  have  the  epoch  e  =~  radians,  or  one  fourth  of  the  periodic 
time.    This  gives  a;  =  r  sin  oot. 

Since  <»  =  ^,  where  T  is  the  periodic  time,  we  have  for  *  =  0, 
X  =  0;  for  t=^\T,x  =  r;  for  f  =  iT,  a;  =  0;  for  t  =  iT,x  =  -r;  for 
t=T,x  =  0. 


The  curve  is  the  curve  of  sines,  or  the  curve  which  would  be  de- 
scribed by  the  point  P  (page  103)  if,  whUe  M  maintained  its  uniforna 
circular  motion,  the  circle  itself  were  to  move  with  uniform  speed 
in  a  direction  perpendicular  to  CA. 


136 


KINEMATICS   0^   A   POINT — TRANSLATION.       [CHAP.  IV. 


It  is  the  simplest  possible  form  assumed  by  a  vibrating  stringy 
when  it  is  assumed  that  at  each  instant  the  motion  of  every  particle 
of  the  string  is  simple  harmonic. 

If  the  rectilinear  harmonic  motion  is  compound,  we  have  (page 
132)  in  general 

X  =  r  cos  {cot  ±  e)  -\-  Vi  cos  {ooit  ±  ei). 

If  the  displacement  of  one  of  the  motions  is  zero  when  ^  =  0,  we 

have  e  =  „-;  if  ei  =  e  +  nn,  we  have 

X  =  r  sin  oot  +  Vi  sin  {wit  +  nit). 

If  the  period  of  one  motion  is  twice  that  of  the  other,  for  instance, 
we  have  ooi  =  2(»,  and 

X  =  r  sin  {oot)  +  Vi  sin  {2Got  +  nn). 

If  the  difference  of  phase  is  zero,  n  =  0;  and  if  the  amplitudes 
are  equal  also,  we  have 

x  =  r  sin  cot  +  r  sin  {2Got). 


This  gives  a  curve  as  shown  in  the  figure.  ' 
Periods  Unequal. — We  have  in  general 

x  =  r  cos  {oot  +  e),     y  =  ri  cos  {ooit  +  ei) 

for  the  two  component  rectilinear  harmonic  motions  at  right  angles. 
The  elimination  of  t  in  any  case  gives  the  curve  of  resultant  com- 
pound harmonic  motion. 

If  the  periods  of  the  components  are  as  1  to  2,  and  e  is  the 
difference  of  the  epochs,  we  have  for  equal  amplitudes 


X  =  r  cos  {2oot  +  e),    y  =  r  cos  oot. 


Eliminating  t. 


^  =  ^|(?^-l)cose  +  2|^/l-^.sine[, 

which  is  the  general  equation  of  the  curve  for  any  value  of  e. 
Thus  for  e  =  0,  or  equal  epochs, 


or    2/ 


(x  +  A 


which  is  the  equation  of  a  parabola.     For  e 


2' 


=  2^i/l 


y" 


or    r'aP^  —  4y''{r'  —  y\ 


which  is  also  the  equation  of  a  parabola. 


CHAP.  IV.]    TKANSLATIOX — COMPOUND  HARMOXIC  MOTIOif. 


137 


If  we  make  ein  succession,  0, 1,  2,  etc.,  eighths  of  a  circumference, 
we  obtain  a  series  of  curves  as  shown. 


Period.  1^2 


In  the  same  way  we  can  find  the  curve  for  any  ratio  of  periods 
and  difference  of  epoch.  Thus  if  the  periods  are  as  1  to  3  or  2  to  3, 
and  we  make  e  in  succession  0,  1,  2,  etc.,  eighths  of  a  circiunference, 
we  obtain  the  following  series  of  curves : 


£=Jf5r 


--}i7r 


e-^TT 


Period  1 : 3 


Period  2:3 


Blackburn's  Pendulum.  —  The  motion  of  a  pendulum  which 
swings  through  a  small  arc  is,  as  we  shall  see  hereafter  (page  154), 
simple  harmonic,  and  the  projection  of  the  bob  on 
a  horizontal  plane  moves  with  simple  rectilinear 
harmonic  motion. 

Curves  similar  to  those  just  given  are  therefore 
traced  by  Blackburn's  pendulum.  This  consists  of 
two  pendulums,  CED  and  EB,  arranged  so  as  to 
swing  in  two  planes  at  right  angles. 

Any  difference  of  period  may  be  made  by  ad- 
justing the  lengths  of  the  pendulums,  and  they 
may  be  started  with  any  difference  of  epoch.  If 
the  bob  B  is  made  to  trace  its  path  on  a  horizontal 
plane,  we  have,  approximately,  the  compound 
harmonic  curve. 

[Application  of  the  Calculus  to  Harmonic   Motion  — Let  a'  be  the  known 
acceleration  at  the  known  distance  ?■'.     Then  the  acceleration  at  any  other  dis- 
tance is  —r,  where  r  is  the  distance  from  the  centre. 
r' 

For  the  acceleration  in  the  direction  of  the  axis  of  X  we  have  then 


(f  a;  _      a' 

If  "  ~?'^'     '     ' 

and  in  the  direction  of  the  axis  of  Y, 

(Py  _       a 

the  minus  sign  denoting  direction  towards  the  centre. 


(1) 


(3> 


138  KINEMATICS  OF   A.  POINT — TRANSLATION.       [CHAP.  IV. 

Multiply  (1)  by  da;  and  we  have 


d^x  ,  a' 

Integrating, 


,,„  dx= jzdas. 

dv  r 


lf^y=-^,a,^  + Const. 
2\dt  j  2r'     ^ 

Let  — -  =  0,  when  x  =  r,  then  Const.  =  — -r^  and 
dt  2r 

(^y=^Xr^-x^),   or    dti/^  =  - 
\dt)       r  y   r 


dx 


\/r-i  — , 


(3) 


where  we  take  the  minus  sign  to  indicate  that  x  diminishes  as  t  increases,  or 
motion  towards  the  centre. 

In  the  same  way  we  obtain  froitn  (2),  if  —  =  0,  when  y  z=.  n, 

m^:,  =  --^=-_, (4) 

where  we  take  the  plus  sign  to  indicate  that  y  increases  as  t  increases. 
Integrating  (3)  and  (4),  we  have 


ty%-\-G=cos-^-,     or    x  =  r  cos<ty%  +  C  Y, 
f/^+Ci  =sin-i^.    or    y  =  r,sm\ty^  +  Cil, 


(5) 
(6) 


where  C  and  d  are  constants  of  integration. 
Equations  (5)  and  (6)  may  be  written 

«  =  r  cos  C  cos  <  4/  —  —  r  sin  C  sin  <  |/  -  =  ^1  cos  <  |/  ^  -j-  ^a  sin  <  |/^» 

y  =  ri  cos  Cisinti/  —  -\- Ti  sin  d  cos  t y  —  =  Bi  sin  ty  %  +  Bacost  y—, 

where  Ai  =  r  cos  C,   A-t  =  —  r  sin  C,    Bi  =  Vi  cos  Ci,    Bi  =  ri  sin  Ci. 

If  we  find  from  these  equations  the  values  of  sin  and  cos  in  terms  of  x  and 
y  and  add  their  squares,  we  have,  by  eliminating  t, 

x\B:'  +  ^,*)  +  2/^(^.2  +  A^^)  -  2xy{A,Ba  +  A^B,)  =  {A,B,  -  A^B^f  .     (7) 

This  is  the  equation  of  an  ellipse  referred  to  its  centre  and  rectangular 

axes. 

If  we  take  one  of  the  principal  axes  corresponding  with  the  axis  of  X,  and 

dx 
count  the  time  from  the  end  of  this  axis,  we  have  for  t  =  0,  y  =  0  and  — -  =  0 

and  X  =  r.      Therefore,  from  (5)  and  (6),   (7=0  and  Ci  =  0,  and  therefore 
Ai  =  0,   Ai  =  r,   Bi  =  0,  Bi  =  Vi,  and  (7)  becomes 

or  the  equation  of  an  ellipse  referred  to  its  principal  axes. 
We  have  also,  from  (5)  and  (6), 


X  =  r  cos 


y  =  Ti  sm 


/a'         ,     dx  /a'    .         /a' 


CHAP.  IV.]  TKANSLATION— PLANETARY   MOTION.  139 

Therefore  elliptic  harmonic  motion  can  be  considered  as  the  resultant  of  two 
simple  rectilinear  harmonic  motions  at  right  angles  of  the  same  period  and 
different  amplitudes,  so  related  that  the  velocity  of  one  is  zero  when  the  ve- 
locity of  the  other  is  a  maximum,  i.e.,  one  is  at  the  centre  when  the  other  is  at 
its  greatest  range. 

They  therefore  differ  in  epoch  by  90°. 

The  time  of  a  complete  oscillation  is  for  each  of  these  component  motions 

2n  y  --,  and  this  therefore  is  the  periodic  time  in  the  ellipse. 

Planeiary  Motion.— Velocity  Inclined  to  "the  Central  Accelera- 
tion—Acceleration Inversely  as  the  Square  of  the  Radius  Vector— 
Hodograph  a  Circle. — Since  the  acceleration  is  central,   we  have 

(page  130)  r'fij  =  c,  or  co  —  -^,  where  c  is  a  constant  and  eqticd  to 

twice  the  areal  velocity  of  the  radius  vector.      Also  by  assumption 
we  have 

where  a'  is  the  acceleration  at  a  known  distance  /. 

Let  P  be  a  point  which  has  the  ve-  ^v 

locity  V,  and  central  acceleration  direct-  ^ — 

ed  always  towards  the    point  O,  the  ^,„^<^^^^^ 

radius  vector  being  OP  =  r.  ^"''^V 

Take  O'  as  the  pole  of  the  hodograph  y^  V 

(page  52),  and  draw  OQ   parallel  and  ^         /^ A___b' 

eqvial  to  v.      Then  the  tangent  to  the  /  o 

hodograph  at  Q  is  the  direction  of  the        ' 

acceleration  /  at  P  and  is  parallel  to  ^ 

OP  =  r.  V "^ 

Since  the  angular  velocity  a?  at  P  is  /^  ^^^^ 

the  angular  velocity  of  the  radius  vec-  /     o'^-^^/^  I  \\ 

tor  ?',  the  angular  velocity  of  the  tan-  aI ^^^ — njbV 

gent  at  Q  is  also  oo.  \       ^  j 

Let  C  be  the  centre  of  curvature  of  V  / 

the  hodograph,  so  that  CQ  is  perpendic-  X^^    ^^^ 

ular  to  the  tangent  at  Q  and  CQ  =  p  = 
radius  of  curvature.    Then  since  the  linear  acceleration  f  ot  P  is 

f 
the  linear  velocity  of  Q,  we  have  f  =  poo,  or  p  =  —. 

c  a!  r'^  ct'/^^ 

But  00=  -  and  /  =  — ^ .     Hence  p  = .        The   radius   of 

r^  r^  c 

curvature  p  is  therefore  constant  and  the  hodograph  for  planetary 

motion  is  a  circle. 

The  path  which,  in  consequence  of  aberration,  a  fixed  star  seems  to  de- 
scribe is  the  hodograph  of  the  earth's  orbit,  and  is  therefore  a  circle  whose 
plane  is  parallel  to  the  plane  of  the  ecliptic. 

The  Path  for  Planetary  Motion  is  a  Conic  Section.  —  Draw  OR 
perpendicular  to  CQ  and  therefore  parallel  to  r.  OP  is  the  com- 
ponent of  the  velocity  v  in  the  direction  of  the  radius  vector.  Draw 
QN  perpendicular  to  OCB.  Then^iV  is  the  component  of  v  per- 
pendicular to  the  fixed  line  CB. 

But  by  similar  triangles 

OR       QN  OR       O'C  .      . 

ac  =  CQ^  ^"  :p^  =  W^'^^''^'''^''*' 


140  KINEMATICS    OF   A    POINT — TEANSLATION.       [CHAP.  IV. 

that  is,  the  ratio  of  the  velocity  along  the  radius  vector  to  th& 
velocity  at  right  angles  to  any  fixed  line  parallel  to  (yCB  is  con- 
stant and  equal  to  e. 

If  then  Vi  and  r-i  are  the  initial  and  final  values  of  r  for  an  in- 
definitely short  time,  and  di ,  d^  the  corresponding  distances  of  P 
from  any  given  fixed  line  A'B'  parallel  to  O'CB,  we  have 

-QN^d7^^.='^    ^^   n-r.  =  e((i.-d.).  .    .    .    (1) 
Since  this  holds  whenever  we  take  the  fixed  line  A'B\  let  u» 
take  the  initial  distance  di  such  that  di  =  -  ,  or  e  =  — -. 

Then,  from  (1),  d^  =  — ,  or  e  =  -^,  and  we  have 
e  di 

II.  —  Il_  — 
d\        di 

that  is,  the  ratio  of  the  distance  r  of  the  moving  point  P  from  a 
fixed  point  O  to  its  distance  d  from  a  fixed  line  has  a  constant 
value. 

This  is  the  Property  of  a  Point  Moving  in  a  Conic  Section.  If 
e  =  1,  then  O'R  =  QN.  and  the  pole  O'  is  on  the  circumference  of  the 
hodograph,  and  the  path  of  P  is  a  parabola. 

If  e  is  less  than  unity,  the  pole  O'  is  inside  the  hodograph  and 
the  path  of  P  is  an  ellipse. 

If  e  is  greater  than  unity,  the  pole  O'  is  outside  the  hodograph 
and  the  path  of  P  is  an  hyperbola. 

When,  therefore,  a  point  has  a  central  acceleration  inversely 
proportional  to  the  square  of  the  distance  from  the  centre,  it  must 
move  in  a  conic  sectio7i  with  the  centre  of  acceleration  as  a  focus. 

Conversely,  if  the  path  be  a  conic  section  and  the  acceleration  is 
directed  towards  either  focus,  it  must  vary  inversely  as  the  square 
of  the  distance  from  the  focus. 

In  both  cases  the  radius  vector  describes  equal  areas  in  equal 
times  (page  130). 

Kepler^s  Laws. — By  laborious  comparison  of  the  observations 
which  Tycho  Brahe  had  made  through  many  years  of  the  planets, 
especially  of  Mars,  Kepler  discovered  the  three  laws  of  planetary 
motion  which  are  known  as  Kepler's  Laws.  He  gave  these  laws 
simply  as  the  expression  of  facts  which  seemed  warranted  by  the 
observations. 

The  three  laws  are  as  follows  : 

I.  The  planets  describe  ellipses,  the  sun  occupying  one  of  the 
foci. 

II.  The  radius  vector  of  each  plnnet  describes  equal  areas  in 
equal  times. 

III.  The  "Harmonic  Law,"  so  called.  The  squares  of  the  period.^ 
of  the  planets  are  proportional  to  the  cubes  of  their  mean  distances, 
from  the  sun. 

The  second  law,  as  we  have  seen  (page  130),  is  a  necessary  con- 
sequence of  central  acceleration. 

From  the  first  law,  as  we  have  just  seen,  it  follows  that  the 
acceleration  must  be  inversely  as  the  square  of  the  distance. 

The  third  law  is  a  direct  consequence  of  such  central  accelera- 
tion, as  we  shall  see  in  the  next  Article. 

Verification  by  Application  to  the  Moon. — Assuming  Kepler's 
third  law,  Newton  was  led  directly  to  the  conclusion  that  the  ao- 


CHAP.  IV.]  TRAIfSLATION — PLANETARY    MOTION.  141 

•celeration  must  be  inversely  as  the  square  of  the  distance,  as 
follows : 

The  moon  and  other  satellites  move  around  their  primaries  in 
sensibly  circular  orbits,  the  centre  being  at  the  centre  of  the 
primary. 

If  T  and  Ti  are  the  periodic  times  of  two  satellites,  then  accord- 
ing to  Kepler's  third  law,  if  r  and  r.  are  the  radii  of  the  orbits, 
we  must  have 

If  00  is  the  angular  velocity  of  one  satellite,  we  have  (page  76) 
rGo  =  -=-,  or    Gj  =  — .    We  have  also  for  the  acceleration  (page 

77)  /  =  -  =  raj*  =  -— -.    For  the  other  satellite  we  have  in  like 

manner  /i  =  -?jrr-    ^®  have  then 

/.  ^  rv,  _  r' 

/       TiV  ~  ri"' 

or  the  acceleration  is  inversely  as  the  square  of  the  distance.  Con- 
versely, if  the  acceleration  is  inversely  as  the  square  of  the  distance, 
Kepler's  third  law  is  a  necessary  consequence. 

The  numerical  verification  of  this  conclusion  by  the  moon  is 
given  in  Example  17  (page  55). 

[Application  of  the  Calculus  to  Planetary  Motion.] — The  general 
formulas  for  central  acceleration  have  been  already  given,  Chap.  VIII,  page 
85. 

For  any  given  law  of  central  acceleration  we  have  only  to  insert  the  cor- 
responding value  oijfdr  in  these  general  equations. 

(a)  To  Determine  the  Path  when  the  Central  Acceleration  'Varies  Inversely 
as  the  Sqnare  of  the  Distance  from  the  Pole. — When  the  acceleration  is  inversely 
as  the  square  of  the  distance  we  have 

where  a'  is  the  acceleration  at  a  known  distance  r',  and  therefore  aV*  is  con- 
stant.    We  have  in  this  case 


//*=/ 


a'r'^dr  _      a'r'' 
~        r 


Substituting  this  in  equation  (42),  page  87,  we  have  for  the  differential  equa- 
tion of  the  path 

/  <fr*     .    1\  ,  2a' r'^ 


ov,  taking  clB  as  always  positive, 

dr 
dS 


„    A.       2aV^       1 


C 


If  we  put  z=- —  a'r"^,  and  »*  =  a'^r*  +  Cic^  this  becomes 
d9  = , 


142  KINEMATICS   OF  A   POINT — TEANSLATION".       [CHAP.  IV. 

Integrating,  we  have 

9  =  cos'^  — \-  Const, 


n 


W nen  z  =  n  and  tnerefore  —  = ^ —  =  — j-  +  4/ — ,  let  9  = 

0.    Then  Const.  =  (p  and 

_i   z 

Q  —  d>  —  cos      — ,     or    s  =  «,  cos  (S  —  d)). 
Replacing  the  values  of  s  and  n,  we  have  for  the  polar  equation  of  the  path 

(1) 


l  +  |/l+^cos(e-0) 

This  is  the  polar  equation  of  a  conic  section  with  the  pole  at  a  focus.     It 

< 
will  be  an  ellipse,  parabola  or  hyperbola  according  as  Ci  =  0. 

fdr  = ,  if  Ti  and  Vi  are  the 

initial  distance  and  velocity  at  any  instant, 

Ci  =  Vi^ , (2) 

< 
and  according  as  this  is  =  0  we  have  an  ellipse,  parabola  or  hyperbola.     We 

> 
see  then  that  the  form  of  the  orbit  depends  solely  upon  the  magnitude  of  the  initial 
velocity  and  not  upon  its  direction.     Also  if  a'r'^  is  negative,  that  is,  when  the 
acceleration  is  directed  away  from  the  pole,  we  have  always  an  hyperbola. 

From  page  100  we  have  seen  that  the  speed  attained  by  a  body  starting  from 
rest  at  an  infinite  distance  from  the  centre  and  moving  in  a  straight  line  with 

/2a'r'^ 

an  acceleration  inversely  as  the  square  of  the  distance  is  4/  . 

^      ri 
Hence  the  orbit  will  be  an  ellipse,  parabola  or  hyperbola  according  as  the 
velocity  of  projection  is  less  than,  equal  to  or  greater  than  that  acquired  from  an 
infinite  distance. 

If  ei  is  the  angle  of  ®i  with  ri ,  we  have  from  equation  (30),  page  85, 

c  =  ri«i  sin  ei (3) 

The  constants  are  therefore  given  by  (2)  and  (3)  and  the  orbit  is  determined 
when  the  initial  velocity  «i  at  the  distance  ri  and  with  the  direction  6i  are 
given. 

{b)  Central  Acceleration  Inversely  as  the  Square  of  the  Distance  from  the 
FocuB — Path  an  Ellipse. — When  the  path  is  an  ellipse  we  have  the  case  of  plan- 
etary motion. 

Let  the  point  Pmove  in  the  ellipse  ABA'  with  central  acceleration  always 

P  directed  towards  the  sun  8  in  the  focus,  and 

B^-'^^    IP"^A  varying  inversely  as  the  square  of  the  dis- 

tance or  radius  vector  8P  =  r. 
'  X        Let  the  semi-major  axis  OA  =  A  and  the 
semi-conj  ugate  axis  OB  =  B  and  the  eccen- 

.  ■  :     08 
tncity  -J-  =  e. 

Let  the  angle  ASX  of  the  major  axis 
•with  any  fixed  line  SX  through  the  focus  be  0. 


CHAP.  IV.]  TRANSLATION — PLANETARY   MOTION.  143 

The  extremity  A  of  the  major  axis  nearest  to  the  focus  S  is  called  in  general 
the  lower  apsis,  or,  in  the  case  of  a  planet,  the  perihelion.  The  angle  ASX  is 
the  longitude  of  the  perilielion.  The  distance  SA  is  the  lower  apsidal  distance, 
or  the  perihelion  distance.  The  other  extremity  A'  is  the  higher  apsis,  or,  in  the 
case  of  a  planet,  the  aphelion,  and  8 A'  is  the  higher  apsidal  distance,  or  aphelion 
distance. 

The  angle  PSX,  or  the  angle  of  the  radius  vector  with  the  fixed  line  8X,  we 
denote  by  6. 

The  angle  PSA  made  by  the  radius  vector  with  the  major  axis  is  then  Q  —  <p. 
This  angle  is  called  the  true  anomaly. 

The  polar  equation  of  an  ellipse  with  reference  to  the  focus  3  as  &  pole, 
counting  the  angle  PSA  around  from  the  perihelion,  is 

l  +  ecos(a-0)  ^^ 

But  equation  (1),  page  142,  just  deduced,  is  the  equation  of  a  conic  section, 
which  becomes  an  ellipse,  therefore,  when 

es^l  +  ^l^      and    ^(1  -  e«)  = -^,    or    A  =  -^^. 
a^r*  ar  Ci 

Inserting  the  values  of  (2)  and  (3),  we  have 

(2a'r'*\ 

e-  =  l-^ :y^. --^ (5) 


ar  -"ri 


2fflV2-ri«,*' 


(6) 


where  Vi  is  the  initial  velocity  at  the  distance  Ti  making  the  angle  ei  with  ri. 

The  elliptic  orbit  is  thus  determined. 

From  (6)  we  see  that  the  semi-major  axis  A  depends  only  on  the  distance  Ti 
and  the  velocity  of  projection  Vi  and  isindependentof  the  direction  of  projection 
€i.  In  whatever  direction  the  body  is  projected,  the  major  axis  will  be  the 
same  for  the  same  distance  and  velocity  of  projection. 

We  have  also 

«'^'-77T^^ (^> 

A{1  —  r) 

But  we  know  ,'page  85)  that  c  is  twice  the  areal  velocity  of  the  radius 
vector.     If  T  is  the  periodic  time,  then,  since  twice  the  area  of  an  ellipse  is 

27CA^  4/1  —  6*,  we  have 

cT  =  2itAW^-e\ 

or 

c2  ,  ,-      47r2^» 


Aa-e')-""    ~     T'    ' 
But  by  KBPliER's  third  law  we  have  for  two  different  planets 
'p  _  A^         A^  _Ai^ 
T?  ^  A?'  ^^  T*  ~  Ti^' 

Hence ,  or  a'r"  is  constant  for  aU  the  planets. 

^(1— e**) 

But  «'  is  the  acceleration  at  a  distance  r',  and  aV^  is  equal  in  magnitude  to 
the  acceleration  at  the  distance  unity,  since  the  acceleration  at  any  distance  r 

.    „      «V^ 

The  direct  consequence  of  Kepler's  third  law,  therefore,  is  that /or  aU  the 
planeU  the  acceleration  is  tM  same  at  the  same  distance  from  the  sun. 


144  KINEMATICS   OF   A   POINT — TRANSLATION.       [CHAP.  IV. 

•'  Of  all  the  laws,"  says  Sir  John  Herschel,  "  to  which  induction  from  pure 
observation  has  ever  conducted  man,  this  third  law  of  Kepler  may  justly  be 
regarded  as  the  most  remarkable,  and  the  most  pregnant  with  important  conse- 
quences. When  we  contemplate  the  constituents  of  the  planetary  system  from 
the  point  of  view  which  this  relation  affords  us,  it  is  no  longer  mere  analogy 
which  strikes  us,  no  longer  a  general  resemblance  among  them  as  individuals 
independent  of  each  other,  and  circulating  about  the  sun,  each  according  to  its 
own  peculiar  nature,  and  connected  with  it  by  its  own  peculiar  tie.  The  re- 
semblance is  now  perceived  to  be  a  trxie  family  likeness  ;  they  are  bound  up  in 
one  chain  ;  interwoven  in  one  web  of  mutual  relation  and  harmonious  agree- 
ment ;  subjected  to  one  pervadmg  influence,  which  extends  from  the  centre  to 
the  farthest  limit  of  that  great  system,  of  which  all  of  them,  the  earth  included, 
must  henceforth  be  regarded  as  members."  * 

(c)  Value  of  a'  for  Planetary  Motion. — In  all  our  equations  for 
central  acceleration  we  see  that  it  is  necessary  to  know  the  accel- 
eration a'  at  some  known  distance  r'. 

We  are  dealing  in  this  portion  of  our  subject  with  Kinematics,  or 
the  study  of  motion  only,  apart  from  its  causes  or  the  properties  of 
matter.  It  is  therefore  not  the  place  here  to  deal  with  ideas  of 
"  force  "  and  "  mass." 

It  is  sufficient  to  state  here  that  the  "mass"  of  a  body  is  the 
number  of  standard  pounds  it  will  balance  at  any  point  of  the 
earth's  surface  in  an  equal-armed  balance.  The  unit  of  mass  is 
then  the  pound. 

It  will  be  proved  hereafter  (see  Vol.  2,  Statics)  that  if  M  is  the 
mass  of  the  sun,  and  m  the  mass  of  a  planet,  the  value  of  a'  which 
must  be  used  in  all  equations  for  planetary  motion  is  given  by 

a  =  .    9, (1) 

m 

where  m'  is  the  mass  of  the  earth,  and  g  the  mean  acceleration  of  a 
body  at  the  earth's  surface  due  to  gravity. 

If  the  two  attracting  bodies  are  the  earth  and  a  small  body  of 

mass  m,  then  a'  =  — —g,  or,  since  m  is  insignificant  with  respect 

to  m',  a'  =  g. 

If  in  the  preceding  Article  we  had  used  the  value  of  a'  given  by 
<1),  we  should  have  obtained 

M+m     ,,      47C^A*         ,    M+m,      ,„      4;r»^i' 
-m^9r^  =  -f^   and    —-^gr^=-^, 

and  hence 


A'  Ai' 


M+m  '  M  +  m,i' 

We  see  then  that  Kepler's  third  law  is  not,  strictly  speaking,  ex- 

a'v'^ 
act.    The  value  of  — ^ ,  or  the  acceleration  at  the  same  distance,  is 

not,  strictly  speaking,  constant  for  all  the  planets.  The  more  accu- 
rate expression  of  the  third  law  is  that  the  squares  of  the  periodic 
times  are  directly  as  the  cubes  of  the  semi-major  axes  and  inverse- 
ly as  the  sum  of  the  mnsses  of  the  sun  and  planet. 

The  error  from  this  source  is,  however,  too  slight  to  be  percept- 
ible, the  mass  of  Jupiter,  the  largest  of  the  planets,  being  less  than 
a  thousandth  part  of  that  of  the  sun. 

*  Outlines  of  Astronomy. 


CHAP.  IV.]  TKANSLATION — PLANETARY   MOTION.  145 

The  motion  of  translation  of  the  planets  is  not  affected  by  their 
rotation  on  their  axes,  and  we  may  treat  them,  therefore,  as  mate- 
rial points  so  far  as  translation  is  concerned. 

The  sun  is  not,  strictly  speaking,  a  fixed  point  in  this  sense,  but 
both  sun  and  planet  move  in  orbits,  so  that  the  pole  or  focus  is  not  at 
the  sun's  centre,  and  this  affects  the  accuracy  of  Kepler's  first  two 
laws.  The  sun  is  also  attracted  by  the  other  planets,  and  the  plan- 
ets attract  each  other. 

The  attraction  of  the  planets  for  each  other  sensibly  modifies 
their  orbits.  The  ellipse  is  therefore  only  an  approximation  to  the 
path,  and  requires  correction. 

Kepler's  laws  are  thus  only  approximate  expressions.  If  there 
were  but  two  bodies,  one  fixed  and  the  other  free  to  move,  then 
Kepler's  first  two  laws  would  be  accurate,  and  the  third  law  would 
approach  accuracy  as  the  mass  of  the  moving  body  becomes  in- 
significant with  respect  to  the  mass  of  the  other. 

(d)  To  Find  the  Velocity  of  a  Planet  at  any  Point  of  its  Orbit. — ^From  equa- 
tion (37),  page  87,  Chap.  VIII,  we  have 


=  Ci 


iffdv. 


fdv  = ,  where  a'  is  given  in  the 

preceding  Article,  and  r'  is  the  mean  radius  of  the  earth.     Also  from  equation 
<2),  page  142, 

,      2aV* 

where  Vi  is  the  velocity  at  the  distance  ri.     Therefore 

v^=v.^  +  2a'r'^(^^-l^. (1) 

or,  since  (page  143) a'r'^  =    ■,.■  _    j., 

„  o   ,  2c«      /I       1\ 

From  page  85,  'o'  =  — j,  ai^d  for  an  ellipse,  from  Analytical  Geometry, 

n^  —  — -e-j —.      Hence 

^  2A  -  r 

'•  =  ffii§  =  "^— ) <^ 

Equation  (2)  gives  the  velocity  for  any  distance  r  if  the  semi-major  axis  A 
is  known. 

Equation  (1)  becomes  the  same  as  equation  (2),  page  99,  for  rectilinear  mo- 
tion. 

COR.  1.  We  see  that  the  velocity  is  greatest  where  r  is  least,  or  at  pen- 
he!  ion,  and  least  at  aphelion,  where  r  is  greatest. 

Cor.  2.  If  a  point  moves  in  a  circle  of  radius  r  with  a  speed  Vi,  its  central 

acceleration  is  —  (page  53).      If  this  acceleration  is  equal  at  any  mstant  to 

r 
the  acceleration  of  the  planet,  we  have  from  equation  (7),  page  143, 

Vi^  _  a'r'^   _  ^'         •'• 

V  "    r^     ""  ^(1  —  e«)  r^' 


146  KINEMATICS   OF   A   POINT — TRANSLATION.       [CHAP.  IV., 

Therefore,  from  (2), 

.    c\2A-r)    .  c^        1 


"  A\l  -e>    ■  A{1  -e'')r' 
or 

9^  :  Vi''  ::  2A  -  r  :  A. 

That  is,  tTie  squa/re  of  the  speed  in  the  ellipse  is  to  the  squa/re  of  the  speed  in 
the  circle  as  the  distance  of  the  planet  from  the  unoccupied  focus  is  to  the  semi- 
major  axis. 

Cor.  3.  If  rj  is  the  perihelion  distance  and  Ti  the  aphelion  distance,  we 
have,  from  (2), 

for  r  =  ri ,   «*  =  — ; 

A    ri 

a'r^  r, 
for  r  =  ra,   v^  =  — ^ ; 

A    Ta 

■while  for  r  =  Awe  have 

That  is,  the  speed  at  tTie  extremity  of  the  minor  aads  is  a  mean  proportional 
between  th^  speeds  at  perihelion  and  aphelton. 

{e)  To  Find  the  Time  of  Describing  any  Portion  of  a  Planet's  Orbit. — From. 

/a'r''^ 
fdr  = , 

W^r^         r     ""'• 

From  page  143  we  have  c^  =  a'r'^A{l  —  e^)    and    Ci  = —. 

A 

r^A 
Substituting  these  values  and  multiplying  by  —r-^,  we  obtain 

Ar^dr^ 

Hence 

at  =  M  \i  rdr   ^^^^ 

[aVy    |/4V  -{A-  rf 
In  order  to  integrate  this  expression,  let  A  —  r  =  Aez,  then 

_  /  ^Mi(l  -  ez)dz  _  I  ^U  j      -  <fe       ■        ezdz       \ 

^*-     \a'r^)    |/nr^  "WW   (  Vi^^     VT^^  r 

Integrating, 

t  =  f-^^*] cos-la  -e{l  -  z^)^\+  Const. 
\a'r'^  J 

When  z  =  1,  or  r  =  J.  —  Ae,  or  the  planet  is  at  perihelion,  let  t  =  ti. 

Then  Const.  =  ti  =  the  time  at  perihelion. 

Hence  the  time  for  any  portion  of  the  path  from  perihelion  is 


t-ti  =  l^;Y\  cos-ig  -  e{l  -  z'')^}, 

\ar  *  / 


(1> 


A  —  T 

■where  z  =  — ^ — ,  or  r  =  All  —  ez). 
Ae 

When  2  =  — 1,  orr  =  ^  +  Ae,  the  planet  is  at  aphelion,  and  t  —  ti  is  the 

time  of  half  a  revolution,  or  , 

■where  T  is  the  periodic  time. 


CHAP.  IV.]  TRANSLATION — PLAKETART  MOTION".  147 

We  have  then 

or  the  square  of  the  periodic  time  varies  as  the  cvbe  of  the  semi-major  axis. 

EXAMPLES. 

(1)  Find  the  speed  and  periodic  time  of  a  body  moving  in  a  circle 
at  a  distance  from  the  earth's  centre  of  n  times  the  earths  radius, 
the  acceleration  being  inversely  as  the  square  of  the  distance. 


--  =  (?)*. --(tT 


(2)  A  body  at  a  distance  ri  from  the  centre  of  the  earth  is  pro- 
jected in  a  direction  which  makes  an  angle  of  60°  with  the  distance  ri , 
with  a  speed  Vi  which  is  to  the  speed  acquired  by  falling  from  an 
infinite  distance  as  1  to  ^B ;  the  acceleration  varying  inversely  as 
the  square  of  the  distance.  Find  the  major  axis,  the  eccentricity, 
the  periodic  time  and  the  position  of  the  lower  apsis. 

Ans    2A  =  -Ti ,    e  =-7r,     T  =     .  /i/ — - ,  where  7*'  is  the  radius  of  the 
2  3  4r    r     g   ' 

earth.     The  lower  apsis  is  at  a  distance  ^i  from  the  focus. 

(3)  A  body  revolves  about  a  centre,  the  acceleration  directed 
towards  the  centre  and  varying  directly  as  the  distance.  To  deter- 
mine the  motion. 

From  the  general  equations  of  page  86  we  can  determine  the  motion.     In 

a'  /*  a'r^ 

the  present  case  we  have  f  =  —^,  and  therefore  /  fdr  =  -^. 

Substituting  this  in  equation  (42),  page  87,  we  have  for  the  differential 
equation  of  the  path 

J  dr^     .    1\  a: 

or,  since  dB  is  always  positive, 

dr 


y  c*       cV      r«* 

If  we  put  g  =  ^  -  ■^,  and  v?  = ^^^ ,  we  have 

-de 


2  Vn^  -  s« 
Integrating,  we  have 

1  s 

9  =  —  cos-i [_  Const. 

3  n 

When  g  =  n,  let  6  =  (p.     Then  Const.  =  <p  and 

1  s 

(e  _  0)  —  -  cos  -1  — ,    or    2  =  n  cos  2(0  —  0). 

2  n 


148  KIlfEMATICS  OF  A  POINT — TRANSLATION.       [CHAP.  IV. 

Replacing  the  value  of  z,  we  have  after  reduction,  for  the  polar  equation 
of  the  path, 

2e^ 


Ci  +  2r«!^ 


*"''-,sin^(O-0) 


Ci  +  inc^ 


This  is  the  polar  equation  of  an  ellipse,  the  pole  at  the  centre  of  the  ellipse, 
where 


Ci  +  2nc^'        B^  ci  +  2nc^ '  Ci  —  2nc^ ' 

For  the  values  of  c  and  Ci  we  have  from  equations  (30)  and  (37),  page  87, 

Ci  =  Vi^  -\ 1—,     and    c  =  ?'ii)i  sin  ei , 

where  Vi  is  the  initial  velocity  of  projection  at  a  given  instant,  Ti  the  corre- 
sponding distance  from  the  centre  at  that  instant,  and  ei  the  angle  of  Vi 
with  Ti. 

The  path  is  therefore  fully  determined. 

To  find  the  periodic  time,  since  the  area  of  the  ellipse  is  tcAB,  and  since  c  is 
twice  the  area  described  in  a  unit  of  time  by  the  radius  vector,  we  have 

cT=27tAB,    or    T=^^^^. 
c 

Inserting  the  values  of  A  and  B,  we  have 

„  4*c 


Inserting  the  value  of  w^  we  have 

2Tt 


/: 


This  is  the  same  result  found  in  page  107  for  rectilinear  harmonic  motion. 

(4)  A  particle  describes  an  ellipse  under  the  action  of  central  ac- 
celeration, directed  to  the  centre  of  the  ellipse.  To  determine  the 
law  of  acceleration. 

The  polar  equation  to  the  ellipse,  centre  pole,  is 

from  which  we  have 

DiflEerentiating,  we  have 


and  hence 

dr"^ 


dr         n         1  \        „    .    „ 


CHAP.  IV.]  TRANSLATION — PLANETARY   MOTION.  149 


Differentiating  again, 


=  [l--J)[^'''-^^'') 


But  from  equation  (49),  page  88,  we  have  cPQ  — .     Therefore 

r 

(Pr     .     dr^  2,1,1 


».2    I      ,4  2  ~r 


Now  from  equation  (45),  page  88,  we  have 

=  cVr-+^i-    LVl-iU^4--l  -4-^"]- 


r  =  -z 7,    or    r  4-  ercosQ  =  ±  Ail  —  e^). 

1  -\-  e  cos  e  '  ^  ' 


The  law  of  acceleration  is  therefore  that  of  the  direct  distance. 

(5)  A  particle  describes  a  conic  section  under  the  action  of  central 
acceleration  directed  to  one  of  the  foci.  To  find  the  law  of  accelera- 
tion. 

The  polar  equation  of  either  the  ellipse  or  hyperbola,  focus  pole,  is 

r 

Differentiating, 

dr  4"  c  cos  Qdr  —  er  sin  BdQ  =  0. 
Differentiating  again, 
d'r  -\-  e  cos  6d^r  —  e  sin  6d9dr  —  e  sin  QdOdr  —  er  cos  QdQ^  —  er  sin  Qd^Q  =  0. 

But  from  eq.  (49),  page  88,  d^B  =  -  ?^^.     Therefore 

JO     .  /I  J  0  /ij«o  ^r         er  cos  0 

d^r  +  e  cos  6dr^  =  er  cos  SdS^,     or 


dB^      1  +  «  cos  9* 
Substituting  this  in  eq.  (45),  page  88,  we  have 

cVl e  cos  9      \—^^  ^ ____£!___ 

•'  ~  r^^l^r        r(l  -f  e  cos  B))  ~  r*  r(l  +  e  cos  9)  ~  ±  A{1  -  e^)r^ 

The  acceleration  is  therefore  inversely  as  the  square  of  the  distance  for  either 
ellipse  or  hyperbola. 

The  polar  equation  of  the  parabola  is 

r  =  :; 7,    or    r  —  r  cos  9  =  2A. 

1  —  cos  9 

Differentiating,  we  have 

dr  —  dr  cos  B  -\-  r  sin  BdB  =  0. 

Differentiating  again, 

dV  —  dV  cos  e  4-  (fr  sin  9<Z9  +  dr  sin  9^9  +  r  cos  9(Z9«  +  r  sin  BcPB  =  0. 

But  from  eq.  (49),  page  88,  d^B  = ;   therefore 

^  „  «  « j/^o  <^^  *■  cos  9 

(f/.  _  (fV  cos  B  =  —  r  cos  9a9*,     or     -—  =  — 


d9»  1  -  cos  9 


150  KINEMATICS   OF   A    POINT — TRANSLATION.       [CHAP.  IV. 

Substituting  this  in  eq.  (45),  page  88,  we  have 

J  —  r\r'^  /•(!  —  cos  B))  ~  r^  r(l  —  cos 


_£__ 
2l^ 


The  acceleration  is  therefore  inversely  as  the  sqtLare  of  the  distance  for  the 
parabola. 

(6)  A  particle  describes  a  hyperbolic  spiral  under  the  action  of 
central  acceleration  directed  to  the  pole.  To  find  the  law  of  accelera- 
tion. 

The  equation  of  the  hyperbolic  spiral,  centre  pole,  is  vB  =  A.  We  have 
then  edr  +  rdb  =  0,  Qd^r  -f-  drdB  +  drdQ  +  rO'Q  =  0. 

From  eq.  (49),  page  88,  <Pe  =  -  ?^^.     Therefore  6^r  =  0,  or  dV  =  0. 

r 
From  eq.  (45),  page  88,  we  have  then 

or  the  acceleration  is  inversely  as  the  cvbe  of  the  distance. 

(7)  A  particle  describes  the  lemniscate  of  Bernoulli  under  central 
acceleration,  the  centre  being  the  node.  To  find  the  law  of  accelera- 
tion. 

The  perpendicular  from  the  node  on  the  tangent  is  ^  =  ±  x-jr .     Hence 

dp_Sr^  dp   _  12A* 

dr  ~  2A''  p^dr  ~    r^    * 

We  have  from  eq.  (48),  page  88,  therefore, 

J  -      ^1     ' 

(8)  A  particle  describes  a  circle  under  central  acceleration  di- 
rected to  a  point  in  the  circumference.    Find  the  law  of  acceleration. 

The  polar  equation  of  the  circle  is  r  =  2J?  cos  0.     Therefore 

dr  =  —2B  sin  6dd,    and    cPr  =  —  rd9^  -  2B  sin  6d^6. 
But  from  eq.  (49),  page  88, 

_„  2drd6      4Ji  sin  6d6^ 

r  r 

^r  =  -rd6^-'-^^^^^=rdS^-'-^^. 
r  r 

and 

^  -     _  §^ 
d6^~'^        r    ' 

Substituting  in  eq.  (45),  page  88,  we  have 

8W 
J  ^  • 


CHAPTEK  V. 

<:!ONSTRAINED  MOTION  OF  A  POINT.    SIMPLE  PENDULUM. 
MOTION  ON  A  CYCLOID.    MISCELLANEOUS  PROBLEMS. 


Motion  on  an  Inclined  Plane  —  Uniform  Acceleration. — Let  a 
point  have  a  uniform  acceleration  /  in  the  direction  AE,  and  let  the 
point  be  constrained  to  move  in  the  straight  line  f^ 
AB  which  makes  the  angle  a  with  the  horizon. 

The  component  of  the  acceleration  in  the  direc- 
tion of  the  motion  is  then  /  sin  a. 

The  motion  along  AB  is  then  rectilinear  motion 
under  uniform  acceleration  /  sin  a,  and  equations 
(1)  to  (6),  page  93,  apply  directly,  if  we  substitute  /  sin  a  in  place 
of  g. 

If  Vi  is  the  initial  velocity  at  A  and  v  is  the  velocity  at  B,  wo 
have  from  (5),  page  93, 

if  —  vi'  =  2/1  sin  a, 

'where  I  is  the  length  of  the  inclined  plane  AB.    But  I  sin  a  =  AE. 

The  speed,  therefore,  gained  in  moving  from  A  to  B  is  equal  to 
iliat  which  would  he  gained  in  falling  through  AE  with  the  uniform 
acceleration  f. 

The  time  in  falling  from  A  to  ^  is  from  (1),  page  93,  f  =  — ^, 


and  in  passing  from  A  to  J5,  f  = 


V—  Vi 


Hence 


/sin  a 

1=1- 
t      AE' 

or  the  times  are  proportional  to  the  distances  I  and  AE. 

The  distance  passed  through  along  AB  is  from  (3),  page  93, 

l  =  Vxt  -V  —/sin  a.  <', 

where  V\  is  the  initial  velocity. 

If  the  point  starts  from  rest,  we  have  for  the  distance  along  AB 

I  =  7:/ sin  a .  f. 
2'' 

Let  AD  be  the  vertical  diameter  of  a  circle  and 
AB  =  I  any  chord.  Join  DB.  Then  we  have  AB 
=  AD  cos  DAB  =  AD  sin  ABC.     If  AB  =  I,  we 

B    have  also  AB  =  ^ff  sin  ABC.     Therefore  AD  = 

\ft^,  or  t  =|/^^.  This  is  independent  of  the  posi- 
tion of  the  chord  AB,  and  therefore  t  is  the  same  for  any  chord 

through  A  or  D. 

•  '  151 


152  KINEMATICS   OF   A    POINT — TRANSLATION.         [CHAP.  V, 

Hence  for  uniform  acceleration  /,  the  time  of  descent  down  alt 
chords  through  the  highest  and  lowest  points  of  a  circle  are  equal. 
This  property  enables  us  to  find  the  line  of  swiftest  descent  to  a 
given  curve  from  any  point  in  the  same 
vertical  plane. 

Thus  if  EG  is  the  curve  and  A  the  pointy 
draw  AO  vertical  or  parallel  to  the  direc- 
tion of/,  and  with  centre  in  AC  describe  a 
circle  passing  through  A  and  tangent  to 
the  curve  EG  at  P. 

Then  AP  is  the  line  of  swiftest  descent 
from  A  to  the  curve  EG.  For  any  other 
point  p  in  EG,  Ap  cuts  the  circle  in  some  point  q,  and  since  the 
time  from  A  to  g  is  equal  to  that  from  A  to  P,  the  time  from  A  to- 
p  is  greater. 

EXAMPLES. 
g  =  32.16  ft.-per-sec.  per  sec.     Friction,  etc.,  disregarded. 

(1)  Find  the  position  of  a  point  on  the  circumference  of  a  vertical 
circle,  in  order  that  the  tim£  of  rectilinear  descent  from  it  to  the 
centre  may  be  the  same  as  the  time  of  descent  to  the  lowest  point. 
Acceleration  due  to  gravity. 

Ans.  30°  from  the  top. 

(2)  TTie  straight  line  down  which  a  particle  will  slide,  under  the 
action  of  gravity,  in  the  shortest  time  from  a  given  point  to  a  given 
circle  in  the  same  vertical  plane,  is  the  line  joining  the  point  to  the 
upper  or  lower  extremity  of  the  vertical  diameter,  according  as  the 
point  is  tvithin  or  without  the  circle. 

(3)  Find  the  line  of  quickest  descent  from  the  focus  to  a  parabola 
whose  axis  is  vertical  and  vertex  upwards,  and  show  that  its  length 
is  equal  to  that  of  the  latus  rectum.  Acceleration  vertical  and 
uniform. 

(4)  Find  the  straight  line  of  swiftest  descent  from  the  focus  of  a 
parabola  to  the  curve  when  the  axis  is  horizontal.  Acceleration 
vertical  and  uniform. 

(5)  The  times  in  which  heavy  particles  slide  from  rest  down  in- 
clined planes  of  equal  height  are  proportional  to  their  lengths. 

(6)  Show  that  if  a  circle  be  drawn  touching  a  horizontal  straight 
line  in  a  point  P  and  a  given  cuirve  in  a  point  Q  below  P,  PQ  is  the 
line  of  swiftest  descent  to  the  curve,  under  constant  vertical  accel- 
eration. 

(7)  Find  the  straight  line  of  quickest  descent  from  a  given  point 
to  a  given  straight  line,  the  point  and  the  line  being  in  the  same 
vertical  plane.     Acceleration  constant  and  vertical. 

Ans.  From  P,  the  given  point,  draw  a  horizontal  line  meeting  the  given 
line  in  C.  Lay  off  along  the  given  line  CD  equal  to  PC.  PD  is  the  required 
line  of  swiftest  descent. 

(8)  A  given  point  P  is  in  the  same  plane  with  a  given  vertical 
circle  and  outside  it,  the  highest  point  Q  of  the  circle  being  below  P. 
Find  the  line  of  slowest  descent  from  P  to  the  circle.  Acceleration 
constant  and  vertical. 

Ans.  Join  PQ  and  produce  it  to  meet  the  circumference  in  B.  PR  is  the 
line  required. 


CHAP,  v.]  TRANSLATION — CONSTRAINED  MOTION.  153 

(9)  A  number  of  heavy  particles  start  ivithout  velocity  from  a 
common  point  and  slide  down  straight  lines  in  various  directions. 
Show  that  the  locus  of  the  points  reached  by  them,  with  a  given  speed 
is  a  horizontal  plane,  and  that  of  the  points  reached  by  them  in  a 
given  time  is  a  sphere  whose  highest  point  is  the  starting-point. 

(10)  The  times  required  by  heavy  particles  to  descend  in  straight 
lines  from  the  highest  point  in  the  circumference  of  a  vertical  circle 
to  alt  other  points  in  the  circumference  are  the  same. 

Also  to  descend  in  straight  lines  to  the  lowest  point  in  the  cir- 
cumference, from  all  other  points  in  the  circumference,  the  times  are 
the  same. 

(11)  If  heavy  particles  slide  down  the  sides  of  a  right-angled 
triangle  ivhose  hypothenu^e  is  vertical,  they  will  acquire  speeds  pro- 
portional to  the  sides. 

(12)  A  point  having  a  constant  acceleration  of  24:  ft.-per-sec.  per 
sec.  is  constrained  to  move  in  a  direction  in  which  its  speed  changes 
in  1  minute  from  10  to  250  yards  per  sec.  Find  the  inclination  of 
its  direction  of  motion  to  that  of  the  given  acceleration. 

Ans.  60°. 

(13)  A  heavy  particle  is  projected  up  an  inclined  plane  whose  in- 
clination to  the  horizon  is  §0".  Find  the  distance  traversed  during 
a  change  of  speed  from  48  to  IQft.  per  sec. 

Ans.  63.68  ft. 

(14)  A  point  has,  when  1  mile  up  an  incline  of  1  in  50  (i.e.,  one 
having  an  inclination  to  the  horizon  of  sin' ^  ~],  an  upward  ve- 
locity of  30  miles  an  hour,  (a)  In  what  tim^  will  it  come  to  a  stand- 
still f  (b)  If  it  afterwards  slides  down,  with  what  speed  will  it  reach 
the  foot  of  the  incline  f 

Ans.  (a)  1  min.  8.4  sec.     (6)  63.7  miles  per  hour. 

(15)  A  body  slides  from  rest  doivn  a  smooth  inclined  plane  and 
then  falls  to  the  ground.  The  length  of  the  plane  is  IS  ft.,  its  incli- 
nation to  the  horizon  30°,  and  the  height  of  its  lowest  point  from,  the 
ground  40  ft.  Find  the  distance  horizontally  from  the  end  of  the 
plane  to  the  point  where  the  body  reaches  the  ground.  [Take  g  =  B2 
ft.-per-sec.  per  sec] 

Ans.  15  V'S'ft.  

Motion  in  a  Curved  Path— TTniform  Acceleration. — Let  ABCD  be 
any  curved  path,  and  Ad  the  direction  of  the  acceleration/.  Any 
very  small  portion  of  the  curve,  AB,  may 
be  considered  as  a  straight  line.  We 
have  then,  as  on  page  151,  the  change  of 
speed  in  moving  from  A  to  B,  the  same 
as  in  moving  from  A  to  6  with  the  con- 
stant acceleration  /.  So  also,  in  moving 
from  B  to  C,  the  change  of  speed  is  the 
same  as  in  moving  from  6  to  c  with  the 
constant  acceleration /. 

Hence  the  change  of  speed  in  traversing  any  portion  ot  the  patn 
AD  is  the  same  as  in  traversing  with  constant  acceleration  /  the 
projection  Ad  of  the  path  on  a  line  in  the  direction  of  the  accelera- 
tion. 


154 


KINEMATICS   OF   A    POINT — TKANSLATION.         [CHAP.  V. 


If  then  Vi  is  the  initial  speed  at  A,  and  vis  the  speed  at  any  point 
J),  we  have 

v'-v:'  =  2f.Ad. 


tion  in  a  Circle — Uniform  Acceleration. — This  is  the  case  of 
the  simple  pendulum,  which  consists  of  a  heavy- 
particle  attached  to  a  fixed  point  by  a  massless  in- 
extensible  string. 

Let  C  be  the  point  of  suspension  and  CA  the 
radius,  and  let  the  acceleration  /  be  uniform  and 
vertical.  For  any  position  of  the  point  i^  the 
angle  ACP=  &,  and  the  acceleration  may  be  re- 
solved into  a  tangential  component  /  sin  6  and 
into  a  normal  component /cos  9. 

The  normal  component  has  no  effect  upon  the 
motion  in  the  curve  at  P. 

If  the  angle  6  is  very  small,  the  arc  will  not 
differ  materially  from  the  sine,  and  we  have  sin  6 


arc  AP 


,  where  I  is  the  length  of  the  radius  CA. 
The  tangential  acceleration  at  the  point  P  is  then  a  = 


/  X  arc  AP 


It  is  therefore  directly  proportional  to  the  displacement  of  P  from 
A,  measured  along  the  path. 

The  motion  of  P  is  thus  simple  harmonic  motion  about  A  as  a 
centre  (page  103). 

The  periodic  time  is  then  (page  104) 


'  =  27t\/- 


displacement 


tangential  acceleration 


=  2;r 


^  9 


or  for  the  simple  pendulum  the  time  of  a  vibration  is  t 

The  periodic  time  is  therefore  for  small  displacements  independ- 
ent of  the  amplitude,  and  therefore  for  small  arcs  the  oscillations 
are  isochronous. 

The  time  of  oscillation  is  usually  taken  as  half  the  periodic  time, 
or  the  time  between  the  instants  at  which  the  pendulum  reaches 
opposite  ends  of  its  swing.  Thus  the  seconds  pendulum  makes  a 
complete  oscillation  in  2  seconds. 

If  S  is  not  very  small  the  time  is  different,  but  the  variation  is 
practically  very  slight.     (See  page  160.) 

CoR.  If  the  velocity  of  P  at  any  instant  is  not  wholly  in  the 
plane  PCA,  it  may  be  resolved  into  two  components,  one  in  the 
plane  PC  A  and  the  other  perpendicular  to  it,  and  both  tangential 
to  a  spherical  surface  Hence,  in  the  case  in  which  0  is  small,  P's 
motion  may  be  resolved  into  two  simple  harmonic  motions  of  the 
same  period;  and  its  motion  is  therefore  (page  135)  elliptic  har- 
monic motion,  the  period  being  the  common  period  of  the  compo- 
nents The  ellipse  described  will  depend  upon  the  amplitude  and 
epoch  of  the  components  and  therefore  upon  the  magnitude  and 
direction  of  the  initial  velocity  of  P. 

If  G  is  not  very  small,  and  the  component  motions  are  of  differ- 
ent amplitudes,  the  periods  will  have  different  values,  and  the 
point  P  describes  curves  similar  to  those  given  on  page  137. 

If  the  component  motions  are  equal  in  amplitude  and  therefore 


CHAP,  v.]  TRANSLATION— CONSTRAINED  MOTION.  155 

in  period  and  differ  in  phase  by  one  quarter  period,  the  point  P 
moves  (page  134)  in  a  circle  about  the  foot  of  the  perpendicular  on 
CA  as  a  centre.  This  is  the  case  of  the  conical  pendulum. 

EXAMPLES. 

(1)  Find  the  time  of  oscillation  of  a  pendulum  10  ft.  long  at  a 
place  at  which  g  =  S2ft.-per-sec.  per  sec. 

Ans.  1.75  sec. 

(2)  Find  the  length  of  the  seconds  pedulum  at  a  place  at  which 
fif  =  31.9. 

Ans.  3.232  ft. 

(3)  Find  the  length  of  the  pendulum  which  makes  24  beats  in  1 
min.  when  g  =  32.2. 

Ans.  20.39  ft. 

\^i(i)A  seconds  pendulum  is  lengthened  1  per  cent.    How  much 
does  it  lose  per  day  f 
Ans.  7  min.  8.8  sec. 

(5)  The  length  of  the  seconds  pendulum  being  99.414  cm.,  find  the 
valine  of  g. 

Ans.  981.17  cm.-per-sec.  per  sec. 

(6)  A  pendulum  37.8  inches  long  makes  182  beats  in  3  min.  Find 
the  value  of  g. 

Ans.  31.78  it.-per-sec.  per  sec. 

\0>ff)lf  two  pendulums  at  the  same  place  make  25  and  30  oscilla- 
tions respectively  in  1  sec,  what  are  their  relative  lengths  f 

Ans.  1.44  to  1. 

(8)  A  pendulum  which  beats  seconds  at  one  place  is  carried  to 
another  where  it  gains  2  sec.  per  day.  Compare  the  values  of  g  at 
the  two  places. 

An^  As  0.999953  to  1. 

^""^(y)  A  pendulum  which  beats  seconds  at  the  sea-level  is  carried  to    /^^  ,.-.«► 
the  top  of  a  mountain  where  it  loses  40.1  sec.  per  day.    ^7"TfiiTli(ifty-  /^^ 
the  value  of  g  to  be  inversely  proportional  to  the  ^t/tm^^rom  the  hs  cO\a^^^K 
centre  of  the  earth,  and  the  sea-level  tobe  4000  miles  from  that  point,    U 
find  the  height  of  the  mountain. 

Ans.  1.86  miles. 

Motion  in  a  Cycloid — Uniform  Acceleration.— A  cycloid  is  the 
curve  traced  by  a  point  in  the  circumference  of  a  circle  which  rolls 
allong  a  straight  line. 


If  the  circle  -EP  rolls  along  the  Hne  AB,  the  point  P  being  origi- 
nally at  A,  the  path  of  P  is  the  cycloid  ACB. 


156  KINEMATICS  OF  A  POINT — TRANSLATION.        [CHAP.  V. 

If  C  is  the  position  of  P  when  the  diameter  of  the  circle  through 
Pis.  perpendicular  to  AB,  the  line  CD  perpendicular  to  AB  is  the 
axis  and  C  is  the  vertex  of  the  cycloid. 

Let  the  uniform  acceleration  /  be  always  parallel  to  DC  and  ver- 
tical. 

Let  the  moving  point  Q  have  at  ^i  a  speed  zero.  Its  speed  at 
Qi  is  then  (page  154; 

v''  =  2f.N,N^. 

Let  t  be  the  time  in  which  the  point  would  with  the  same  ac- 
celeration and  with  initial  speed  zero  move  from  D  to  C     Then 

GD  =  \ft\    Hence 

a 

u^  =  ^  .  N.N.  .  CD  =  \CD{CN  -  CN,). 
t  t 

Now  by  a  property  of  the  cycloid 

4CD .  CN  =  CQ.'    and    400 .  CN^  =  CQ^. 

Hence 

^  =  \{CQ.-  -  CQ^')- 

2CD 

Now  f  =  — —   is  a  constant.     Hence  the  motion  of  Q  in  the 

cycloid  is  simple  harmonic  (page  103),  where  —  =  — ,   a'  being  the 

tangential  acceleration  of  Q  at  the  distance  s'  measured  along  the 
curve.     If  T  is  the  time  of  a  complete  oscillation,  we  have 


T=2ic  i/%  =  2nt  =  2n  i/^. 
y  a'  ^     f 


It  t'  is  the  time  occupied  in  moving  from  Qi  to  C, 

^2CD 


</' 


or  the  time  of  a  pendulum  whose  length  is  2CD,  or  4  times  the 
radius  of  the  generating  circle. 

As  this  involves  only  constant  quantities,  the  time  is  the  same 
whatever  be  the  position  of  the  starting-point  ^i,  or  the  oscillations 
are  isochronous.    Hence  the  cycloid  is  called  a  tautochrone. 

This  result  is  rendered  of  practical 
importance  by  one  of  the  properties  of 
the  cycloid,  viz.,  that  if  a  flexible  and 
inextensible  string  AB  is  fixed  at  the 
end  A  and  wrapped  tightly  round  the 
semi-cycloid  AC,  the  end  B  of  the  string 
as  it  unwinds  will  describe  another  semi- 
cycloid.  If  then  AC  and  AD  are  fixed 
semi -cycloids,  symmetrical  with  refer- 
ence to  the  vertical  AB,  and  AB  is  a 
simple  pendulum,  B  will  describe  a  cy- 
cloid, and  its  oscillations  will  be  isochronous  whatever  their 
extent. 


CHAP,  v.]  TRANSLATION — CONSTRAINED   MOTION. 


157 


[Application  of  the  Calculus. — To  Determine  the  Motion  of  a  Point  Constrained 
to  Move  in  a  Cycloid,  the  Acceleration  being  Constant,  in  the  Direction  of  the 
Axis  and  towards  the  Vertex.] — By  the  application  of  the  general  formulas  of 
page  88  we  can  deduce  the  results  A_  d  B 

already  obtained. 

Let  the  axis  CD  =  2r,  where  r 
is  the  radius  of  the  generating  circle 
DF'C.  Let  the  acceleration  /  act 
downward.  Let  CN  =  y,  NP—  x 
and  the  length  of  arc  CP  =  s.  Let 
the  initial  position  be  P,  at  the 
height  CNi  =  h  above  C,  and  the 
speed  at  Pi  be  «i  =  0. 

We  have  thus  the  case  of  equation  (55),  page  89,  and  obtain  at  once,  since 
/  is  negative  and  Si  =  h, 

«=  4/a/(A  -  y) 

for  the  speed  at  any  point  given  by  CN  —  y.  When  y  =  0,  we  have,  at  the 
lowest  point  G,  v  =  4/2/%,  which  is  the  same  as  that  due  to  the  vertical  height 
h. 

By  the  property  of  the  cycloid  we  have 

«  =  arc  CP=  2  VDG  X  ON  =  2  )/Wy  =  2  chord  CP'. 
Hence 

ds=  ±  dyif  — 

^  y 

We  substitute  the  minus  value  in  equation  (56),  page  89,  because  for  de- 
scent the  arc  decreases  as  the  time  increases.     We  have  then 


dt 


-s/% 


dy 


f  Vhy-y^ 
Integrating,  since  for  <  =  0,  y  =  h,  we  have 


<=|/^.(;r-versin-^^). 


(1) 


For  the  time  of  descent  to  the  lowest  point  where  y  =  0,  or  for  the  time  of 
one  quarter  of  a  complete  oscillation, 


The  periodic  time  is  then 


,  /r       Tt  .  /At 

<  =  ^r  7  =  2-r  7 


^=2* 


or  the  same  as  a  simple  pendulum  (page  154)  whose  length  is  4  times  the 
radius  of  the  generating  circle  DP'  C.  1,  x  xi  • 

The  time  is  independent  of  h  and  is  the  same  no  matter  what  the  posi- 
tion from  which  the  point  begins  to  descend.  The  oscillations  are  therefore 
isochronous  and  hence  the  cvcloid  is  called  the  tautochrone. 

The  reason  of  this  remarkable  property  is  easily  seen  by  considering  the 
tangential  acceleration.  _,„     m. 

In  the  cycloid  the  chord  CP'  is  always  parallel  to  the  tangent  IP.  Ihe 
tangential  acceleration  or  tangential  component  of/ is  then 


J9  CP' 

g=:/sinriy=/sinC7i)P'=/^^- 


/; 


4r 


The  tangential  acceleration  is  therefore  directly  proportional  to  the  distance 
from  the  vertex  measured  along  the  path,  and  the  motion  of  P  is  simple  har- 
monic (page  103). 


158  KINEMATICS   OF   A   POINT — TRANSLATION.         [CHAP.  T^ 

The  periodic  time  is  then  (page  104) 

/         displacement  /  «  i/^ 

— "    Y    tangential  acceleration  ~       \/f*     ~~  f' 

V    J^ 
as  already  found. 

A  1  TT      /T 

If  in  (1)  we  make  y  =  —  =  ^GNi,  we  have  t  =  jr-^/  -;?,  or  half  the  time 
«         «  a  '     J 

from  Pi  to  C.     The  time,  therefore,  in  descending  through  half  the  vertical 
space  to  C  is  half  the  time  of  passing  from  P^  to  G. 

[To  Find  a  Curve  such  that  a  Point  Moving  on  it  under  the  Action  of  Gravity 
will  Pass  from  any  one  Given  Position  to  any  Other  in  Less  Time  than  by  any  Other 
Curve  through  the  Same  Two  Points.] — This  is  the  celebrated  problem  of  the 
' '  curve  of  swiftest  descent  "  first  propounded  by  BemouUi.  The  following  is 
founded  upon  his  original  solution. 

If  the  time  of  descent  through  the  entire  curve  is  a  minimum,  that  through 
any  portion  of  the  curve  is  a  minimum. 

It  is  also  obvious  that  between  any  two  contiguous  equal  values  of  a  con- 
tinuously varying  quantity,  a  maximum  w  minimum  must  lie. 

This  principle  though  simple  is  of  very  great  power,  and  often  enables  us  to 
solve  problems  of  maxima  and  minima,  such  as  require  not  merely  the  pro- 
cesses of  the  Differential  Calculus  but  those  of  the  Calculus  of  Variations. 
The  present  case  is  a  good  example. 

,p  Let,  then,  PQ,  ^i?and  PQ;,  Q^B  be  two  pairs  of 
indefinitely  small  sides  of  a  polygon  such  that  the 
time  of  descending  through  either  pair,  starting  from 
P,  may  be  equal.  Let  QQ'  be  horizontal  and  in- 
definitely small  compared  with  PQ  and  QR.  The 
curve  of  swiftest  descent  must  lie  between  these  paths, 
and  must  possess  any  property  which  they  have  in 
common  Hence  if  we  draw  Qm,  (^n  perpendicular 
to  RQ,  PQ,  and  let  v  be  the  speed  down  PQ  or  PQ^ 
(supposed  uniform)  and  v'  that  down  QR  or  QR,  we 
have  for  the  time  from  P  to  i2  by  either  path 

V   "^  tf  V      '     v'  '  V  v'         * 


or 


Qn  _  Q'm 
V    ~     if    ' 

Now  if  9  be  the  inclination  of  PQ  to  the  horizon,  and  Q*  that  of  QR,  we 
have  ^  =  Q(^  cos  6,  Q'm  =  Q(^  cos  &'.     Hence 


This  is  true  for  any  two  consecutive  elements  of  the  required  curve,  and 
therefore  throughout  the  curve  we  have,  at  any  point,  v  proportional  to  the 
cosine  of  the  angle  which  the  tangent  to  the  curve  at  that  point  makes  with 
the  horizontal.  But  «'  is  proportional  to  the  vertical  distance  h  fallen  through. 

Hence  the  curve  required  is  such  that  the  cosine  of  the  angle  it  makes  with 
the  horizontal  line  through  the  point  of  departure  varies  as  the  square  root  of 
the  distance  from  that  line 

Now  in  the  figure  of  page  157  we  have,  from  the  property  of  a  cycloid. 


cos  CP'N  =  cos  TPN  =  cos  GDP'  =  ^^  =  y' 


DP'  _     /DN 
DG  ~y  DG' 


The  curve  required  is  tJierefoi'e  the  cycloid.     The  cycloid  has  received  on 
account  of  this  property  the  name  of  BrachistocJirone. 


CHAP,   v.]  TRANSLATION — CONSTRAINED   MOTION.  15* 

[To  Determine  the  Uotion  of  a  Point  Constrained  to  Hove  in  a  Circle,  the  Ao- 
celeration  being  Constant  and  Vertical.] — This  is  the  case  of  equation  (55),  pag» 
89.  A        c B 

Let  DN  =  y,  NP  =  x.     Let  the  speed  at  Pi  be 
Di  =  0,  the  distance  DNi  =  h. 

For  the  speed  at  any  other  point  P  we  have       \  ^i 

at  once  from  equation  (55),  page  89,  since  fy  is 
minus  and  «i  =  h. 


V  =  y2f{h  -  y), 
VFhere  y  is  the  distance  DN.     When  y  =  0,  we  have  for  the  lowest  point,  D» 

which  is  the  same  as  that  due  to  the  vertical  height  h. 
From  equation  (56),  page  89,  we  have 

dt  * 


The  equation  of  the  circle  referred  to  2)  as  origin  is 
x^  =  2rx  —  y^, 
where  r  is  the  radius.     Hence 

2r2/-y* 


But  cZ«»  =  dx»  +  ^'^  =  d/(l  +  ^^^) 


•.  <fe=  ± 


2ry-y^' 
rdy 


4/2ry  —  2/* 

We  substitute  the  minus  value  in  equation  (56),  page  89,  because  the  arc- 
decreases  as  the  time  increases,  and  obtain 

,  r  dy  ^ ^_  ^y  (1^ 

VW     VQi-yt^ry-f)  V^f  Vihy-y'>){2^-y) 


If  y  is  small  in  comparison  to  3r,  this  reduces  to 

1  ./r^    dy 

^y  f'  4/M 


dt 


^hy  -  y* 
Integrating,  since  when  i  =  0,y  =  h,we  have 

When  y  =  0, 

It  ,  pr 

This  is  the  time  of  one  quarter  of  a  complete  oscillation.     The  periodic 


time  is  then 


'=2;r|/^. 


This  is  the  same  result  as  on  page  154.  v.         t        n\ 

If,  however,  y  is  not  small  in  comparison  to  3r,  we  have,  from  (1), 

VW    V^y-y''  ^     ^    ^hy-y'^ 


160  KINEMATICS   OF   A   POINT — TEANSLATION.  [CHAP.  V, 

Expanding  [  1  —  5    1        by  the  binomial  theorem,  we  have 

Thus  the  terms  to  be  integrated  are  of  the  form .  the  exponents 

yhy  -  y'^ 
n  being  the  natural  numbers  beginning  with  zero.     Performing  the  integra- 
tions, and  taking  the  limits  y  =^  h  and  ^  =  0,  we  have  for  the  time  of  one 
quarter  oscillation 

This  series  converges  more  rapidly  as  h  becomes  smaller.     If  we  take  only 

the  first  term  we  have,  as  already  found,  t  =  —4/  —. 

Q  •' 

We  can  put  h  =  r  —  r  cos  6  =  Sr  sin^  — -,  where  6  is  the  semi-angle  of  oscil- 

lation  DGPi  ;  or  taking  the  arc  as  equal  to  its  sine,  we  have  for  the  first  two 
terms 


=  yj(+0 


Thus  if  the  point  swings  through  an  arc  of  —   radian  or  5°. 7,  on  each  side 
of  the  vertical,  the  time  of  an  oscillation  is  increased  by  about  ,  or  in  the 

case  of  the  seconds  pendulum  the  time  of  a  beat  is  increased  by  of  a 

second.     For  a  swing  on  either  side  of  the  vertical  of  any  amount  we  have  for 
the  time  of  a  quarter  oscillation 

.       f  i  /r(,  ,   1     .  ,  6    ,    9     .  ^  6    ,    225    .  ,  6  \ 

'  =  ■2^/11+4  "'^'y  +  ei  ""'  2 +2304  ""'  2'  'H 
For  a  swing  of  60°  on  either  side  we  have,  therefore, 


*-  2r /l^+i6  +  1024J' 


or  the  time  is  increased  by  about  t^. 


MISCELLANEOUS   PROBLEMS.  161 


MISCELLANEOUS  PROBLEMS. 

We  give  here  as  exercises  for  the  student  a  number  of  problems  covering 
the  preceding  Chapters.     Resistance  of  the  air  and  friction  are  neglected. 

(1)  A  courier  travels  at  the  rate  of  5^  miles  in  8  hours.  Six  hours 
after  his  departure  another  courier  is  dispatched  from  a  place  12 
miles  behind  the  starting-point  of  the  first.  The  second  courier 
travels  at  the  rate  of  8  miles  in  10  hours  over  the  same  route  as  the 
first.     How  long  before  he  will  overtake  the  first  ? 

Ans.  120  hours. 

(2)  A  man  walks  on  the  deck  of  a  vessel  from  bow  to  stern  at  the 
rate  of  3  miles  an  hour,  ivhile  the  vessel  moves  at  the  rate  of  7  miles 
an  hour.  Find  the  speed  of  the  man  with  reference  to  the  earth's 
surface. 

Ans.  4  miles  an  hour,  in  the  direction  of  the  ship's  motion. 

(3)  A  point  moves  in  a  given  path  for  10  seconds  with  a  uniform 
rate  of  change  of  speed  of  +  8  jt.-per-sec.  per  sec.  Find  the  final 
speed  and  the  space  traversed,  if  the  point  starts  from  rest. 

Ans.  v  =  at,  s  —  -^afi ; 

«  =  80  ft.  per  sec,  s  =  400  ft. 

(4)  A  point  has  an  initial  speed  of  7.7  ft.  per  sec.  and  a  uniform 
rate  of  change  of  speed  of  +  5.5  ft.-per-sec.  per  sec.  Find  the  time 
of  passing  over  2200  ft. 

Ans.  8  —  Vit  +  -^af.     <  =  26.9  sec. 
2 

(5)  A  point  has  an  initial  speed  of  7  ft.  per  sec.  and  a  final  speed 
of  125  ft.  per  sec,  and  describes  a  distance  of  3250  ft.  What  is  the 
uniform  rate  of  change  of  speed  f 

Ans.  s  —  Si  —  — jr ,  a  =  2.4  ft.-per-sec.  per  sec. 

(6)  A  point  has  an  initial  speed  of  100  ft.  per  sec.  and  a  rate  of 
change  of  speed  of  1  ft.-per-sec.  per  sec.  Its  final  speed  is  7  ft.  per 
sec.    Find  the  time  of  motion  and  the  space  described. 

Ans.  V  =  Vi  —  at,  <  =  93  sec. ; 

s  —  Si  =  Vit  —  -^afi,    8  —  8i  =  4975.5  ft. 

(7)  A  locomotive  has  a  speed  of  30  miles  an  hour  on  a  level  when 
the  brakes  are  applied.  The  loss  of  speed  due  to  the  brakes  is  8  ft.- 
per-sec.  per  sec.  Find  (a)  the  speed  at  the  end  of  3  seconds  and  the 
distance  traversed ;  (b)  the  time  of  coming  to  rest;  (c)  the  retarda- 
tion in  order  that  the  locomotive  may  come  to  rest  in  30  seconds. 

Ans.  »  =  «!  —  a<,     ?)i  =  44  ft.  per  sec;     o  =  8  ft.-per-sec.  per  sec; 

a  —  8i  =  Vit  —  -^at^. 

(a)  20  ft.  per  sec,  118  ft.;  (5)  t  =  5.5  sec;  (c)  1.47  ft.-per-sec.  per  sec. 


162  KINEMATICS   OF   A    POINT— TKANSLATION". 

(8)  A  body  falls  for  4  seconds  in  vacuo.  Find  the  final  velocityr 
and  the  space  described  (g  =  32^  ft.-per-sec.  per  sec). 

Ans.  v  =  gt,   8  —  ^gi^; 

V  =  138f  ft.  per  sec,     s  =  257i  ft. 

(9)  A  body  falling  in  vacvo  has  a  final  velocity  of  250  ft.  per  sec. 
Find  the  time  of  falling  from  rest,  and  the  distance  described  {g  = 
32^  ft.-per-sec.  per  sec). 

Ans.  7.77  sec;  971.25  ft. 

(10)  A  body  falling  in  vaxiuofrom  rest  describes  a  distance  of  85 
feet.  Find  the  tims  of  fall  and  the  final  velocity  (g  =  32^  ft.-per- 
sec.  per  sec). 

Ans.  2.3  sec. ;  73.9  ft.  per  sec. 

(11)  A  body  is  projected  vertically  uptvards  in  vacuo  icith  a  ve- 
locity of  40  ft.  per  sec.  Find  the  height  and  the  time  of  ascent  (g  = 
32|  ft.-per-sec.  per  sec). 

Ans.  24.87  ft.;  1.24  sec 

(12)  A  body  projected  vertically  upwards  in  vacuo  returned  after 
18i  seconds.  Find  the  initial  velocity  and  the  height  of  ascent  {g  = 
32^  ft.-per-sec  per  sec). 

Ans.  297.54  ft.  per  sec;  1376  ft. 

(13)  A  body  falling  in  vacuo  has  at  a  given  instant  a  velocity  of 
17  ft.  per  sec,  at  a  later  instant  a  velocity  of  90  ft.  per  sec.  Find 
the  time  between  the  two  instants  and  the  distance  traversed  (g  =  32^ 
ft.-per-sec  per  sec). 

Ans.  2.27  sec;  121.44ft. 

(14)  A  stone  is  dropped  into  a  well  and  the  splash  is  heard  in  3 
seconds.  If  sound  travels  in  air  with  a  uniform  velocity  of  1090  ft. 
per  sec.,  find  the  depth  of  the  well  (g  =  32^  ft.-per-sec  per  sec). 

Ans.  130.4  feet. 

(15)  A  point  has  two  component  velocities  (or  accelerations),  at 
right  angles,  of  35  and  87  units.  Find  the  resultant  velocity  (or  ac- 
cslerations). 

Ans.  93.77  units,  making  an  angle  with  the  35  units  of  68°  5'. 

(16)  A  point  has  a  velocity  of  120  ft.  per  sec.  Resolve  this  into 
two  component  velocities  at  right  angles,  (a)  one  of  the  components 
being  75  ft.  per  sec;  (b)  one  of  the  components  making  an  angle  of 
34°  7'  3"  with  the  resultant. 

Ans.  (n)  93.68  ft.  per  sec,  the  resultant  making  an  angle  with  75  ft.  per 
sec  of  51°  19'  4". 

(b)  99.343  ft.  per  sec.  adjacent  to  the  given  angle  and  67.306  ft.  per  sec. 
opposite. 

(17)  A  point  has  two  accelerations  of  115  and  89  ft.-per-sec  per 
sec.  at  an  angle  of  147°  8'  3  .  (a)  Find  the  resultant  acceleration ;  (b) 
Find  the  angle  between  the  given  accelerations  ichen  the  resultant  is 
equal  to  the  lesser  ;  (c)  when  it  is  equal  to  the  greater. 

Ans.  (a)  62.865  ft.-per-sec.  per  sec,  making  the  angle  of  88°  4'  with  89  ft.- 
per-sec.  per  sec;  (b)  130°  14'  44";  (c)  112°  45'  54'. 

(18)  A  point  has  an  acceleration  of  77.5  ft.-per-sec.  per  sec.  Re- 
solve it  into  two  components  (a)  making  unth  the  given  acceleration 
the  angles  35°  7  11 '  and  52°  9'  8";  (b)  ivhen  one  of  the  components  is 
50.5  ft  -per-sec.  per  sec.  and  makes  an  angle  of  36°  8'  6"  with  the  re- 


MISCELLANEOUS    PROBLEMS.  163 

sultant ;  (c)  when  one  of  the  components  is  60  ft.  -per-sec.  per  sec. 
and  the  other  makes  an  angle  of  47'  10'  11"  with  the  resultant;  (d) 
when  the  two  components  are  46.2  and  35  ft.  -per-sec.  per  sec. 

Ans.  (a)  61.265  and  44.634  ft.  per-sec.  per  sec. 

(6)  47.37  ft.-per-sec.  per  sec,  making  an  angle  witli  the  resultant  of  39°  2' 
8". 

(c)  71.88  or  33.48  ft.-per-sec.  per  sec,  making  an  angle  with  the  resultant 
of  61°  38'  17"  or  34°  9'  4". 

{d)  The  components  make  an  angle  of  35°  4',  and  the  resultant  makes  with 
the  component  46.3  the  angle  15°  3'  18". 

(19)  A  stream  flows  ivith  a  velocity  of  1  ft.  per  sec,  and  a  boat 
whose  speed  is  1.3  ft.  per  sec.  is  steered  up  stream  at  an  angle  of  30° 
witJi  the  current.     Find  the  resultant  velocity. 

Ana.  0.66  ft.  per  sec.  down  stream  at  an  angle  of  79°  3'  with  the  current. 

(20)  What  is  the  ratio  of  the  speed  of  light  to  that  of  a  cannon- 
ball  moving  at  the  rate  of  2400  feet  per  sec,  if  light  passes  from  the 
sun  to  the  earth,  a  distance  of  91,500,000  miles,  in  8^  minutes  f 

Ans.  403600  to  1. 

(21)  ABC  is  a  triangle.  Two  spheres  of  radii  rx  and  r,  start  to- 
gether from  A  and  B,  their  centres  moving  along  AC  and  BC  ivith 
velocities  which  carry  them  separately  to  C  in  the  same  time.  Find 
the  distances  each  has  gone  through  when  they  meet. 

Ans.  If  a,  b  and  c  are  the  sides  of  the  triangle,  the  required  distances  are 

-(c  —  ri  —  Ti),  -(c  -  ri  —  r»). 

(22)  If  a  particle  is  projected  vertically  in  vacuo  with  a  velocity 
of  8g,  find  the  time  in  which  it  will  rise  through  the  height  14gf. 

Ans.  3  sec.  and  14  sec. 

(33)  A  body  falling  in  vacux)  is  observed  to  describe  144. 9  ft.  and 
177. 1  ft.  in  two  successive  seconds.  Find  g  and  the  time  from  the 
beginning  of  the  motion. 

Ans.  g  =  'd2.2  ft.-per-sec  per  sec ;  <  =  4  sec.  to  the  beginning  of  the  first  of 
the  two  seconds. 

(24)  A,  B,  C,  D  are  points  in  a  vertical  line,  the  distances  AB, 
BC,  CD  being  equal.  If  a  particle  fall  from  A,  prove  that  the  times 
of  describing  AB,  BC,  CD  are  «s        _ 

1  :  \/2-  1:VS-  V2~ 

(25)  A  particle  describes  in  successive  intervals  of  4  seconds  each 
spaces  of  24  and  64  feet  in  the  same  straight  line.  Find  the  accelera- 
tion and  the  velocity  at  the  beginning  of  the  first  interval. 

Ans.  3.5  ft.-per-sec.  per  sec;  one  foot  per  sec. 

(26)  A  particle  moves  7  ft.  in  the  first  second,  and  11  and  17  ft. 
in  the  third  and  sixth  seconds  respectively.  Show  that  these  facts 
are  consistent  with  the  supposition  of  a  uniform  acceleration. 

(27)  A  falling  particle  is  observed  at  one  portion  of  its  path  to 
pass  through  nft.  in  s  seconds.  Find  the  distance  described  in  the 
next  s  seconds. 

Ans.  ?i-f-^s'^. 

(28)  If  s,  ms,  are  the  spaces  described  by  a  body  in  times  t,  nt, 
respectively,  determine  the  acceleration  and  the  velocity  of  projection. 

2(m  —  n)s         J     (m  —  n^)s 

Ans.  — ; -T-,    and    —- --. 

n{n  —  \)t^  n{l  —  n)t 


164  KINEMATICS   OF   A   POINT — TRANSLATION. 

(29)  If  the  focus  of  the  path  of  a  projectile  he  as  much  below  the 
horizontal  plane  through  the  point  of  projection  as  the  highest 
point  of  the  path  is  above  it,  to  find  the  angle  of  projection. 

Ans.  If  a'j  is  the  angle  of  projection,  Ui  =  tan -if — ■=-  j. 

(30)  Particles  are  projected  from  the  same  point  in  the  same  di- 
rection with  different  speeds.  Find  the  loctts  of  the  foci  of  their 
joaths. 

Ans.  A  straight  line  through  the  point  of  projection  making  an  angle  with 

It 
the  horizontal  equal  to  —  —  2ai ,  where  ax  is  the  angle  of  projection. 

(31)  If  a  particle  is  projected  in  a  direction  inclined  to  the  hori- 
zon, show  that  the  time  of  moving  between  two  points  at  the  extrem- 
ities of  a  focal  chord  of  the  parabolic  path  is  proportional  to  the 
product  of  the  velocities  of  the  particle  at  the  two  points. 

(32)  Two  particles  are  projected  from  two  given  points  in  the  same 
vertical  line  in  parallel  directions  and  with  equal  speeds.  Prove 
that  tangents  drawn  to  the  path  of  the  lower  will  cut  off  from  the 
path  of  the  upper  arcs  described  in  equal  times. 

(33)  If  a  particle  is  projected  from  a  given  point  so  as  to  strike 
an  inclined  plane  through  that  point  at  right  angles,  prove  that 

tan  {a I  —  B)  =  —cot  a,  where  ai  is  the  angle  which  the  direction  of 

projection  makes  with  the  horizon,  and  Q  is  the  inclitiation  of  the 
plane  to  the  horizon. 

(34)  A  particle  is  projected  from  a  given  point  with  a  given 
velocity.  Find  the  direction  of  projection  in  order  that  its  path 
may  touch  a  given  plane. 

Ans.  Let  ai  be  the  angle  of  projection  with  the  horizontal  and  /i  the  angle 
of  the  given  plane  with  the  horizontal,  •Pj  the  velocity  of  projection  and  d  the 
distance  from  the  point  of  projection  to  where  the  given  olane  cuts  the  hori- 
zontal through  the  point  of  projection.     Then 


cos  (90  —  /3  —  ttj)  =  -^-^ —. 

Vi 

(35)  To  find  the  least  velocity  with  which  a  body  can  be  projected 
from  a  given  point  so  as  to  hit  a  given  mark,  and  the  direction  of 
projection  in  this  case. 

Ans.  Let  d  be  the  horizontal  distance  from  the  point  of  projection  to  the 
mark,  Vi  the  velocity  of  projection,  ai  its  inclination  to  the  horizon  and  /3  the 
angle  of  elevation  of  the  mark  above  the  horizon.     Then 

•■=/^"^^^.  ..=:-(^+4 

(36)  If  two  circles  the  planes  of  which  are  vertical  touch  each 
other  internally  at  their  highest  or  lowest  points,  and  if  any  chord 
be  drawn  within  the  larger  circle,  terminating  respectively  at  its 
highest  or  lowest  point,  prove  that  the  time  of  descent  down  that  por- 
tion of  the  chord  which  is  exterior  to  the  smaller  circle  is  invariable. 

(37)  AP,  PB  are  chords  of  a  circle,  AB  being  the  vertical  di- 
ameter. Particles  starting  simultaneously  from  A,  P,  fall  down 
AP,  PB,  respectively.  Prove  that  the  least  distance  between  them  is 
equal  to  the  distance  PB. 


MISCELLANEOUS   PKOBLEMS.  165 

(38)  Two  circles  lie  in  the  same  plane,  the  lotvest  point  of  one 
being  in  contact  with  the  highest  point  of  the,other.  Prove  that  the 
time  of  descent  from  any  point  of  the  former  to  a  point  of  the  latter 
along  a  straight  line  joining  these  points  and  passing  through  the 
point  of  contact,  is  constant. 

(39)  A  particle  slides  from  rest  down  a  smooth  sloping  roof  and 
then  falls  to  the  ground.  Find  the  point  where  it  reaches  the 
gi'ound. 

Ans.  Let  I  =  the  lengtli  of  the  slope,  a  its  inclination  with  the  horizontal, 
h  =  the  height  of  the  lowest  point  of  the  slope  Irom  the  ground.  Then  the 
distance  of  the  point  where  the  particle  reaches  the  ground  from  the  foot  of 
the  wall  is 

2cosa  i/l  sin  a(  ^l  sin*  a-{-h  —  ^l  sin*  a). 

(40)  Two  equal  inclined  planes  are  placed  JxicTc  to  hack,  and  a 
particle  projected  up  one  flies  over  the  top  and  comes  to  the  ground 
just  at  the  foot  of  the  other.  Find  the  velocity  of  projection,  a  being 
the  inclination  of  each  plane  and  h  their  common  altitude. 

Ans.   Trj/^rA  (8  +  cosec**  ay  • 

(41)  A  particle  is  projected  from  a  point  A  with  the  velocity  ac- 
quired by  falling  down  a  height  a,  up  an  inclined  plane  of  which  the 
base  and  height  are  each  equal  to  b,  and  after  quitting  the  plane 
strikes  the  horizontal  plane  AB  at  the  point  B.    Find  AB. 

Ans.  AB  is  equal  to  a  -|-  {a''  —  6^* . 

(43)  A  particle  slides  doum  a  smooth  inclined  plane.  Determine 
the  point  at  which  the  plane  is  cut  by  the  directrix  of  the  path 
described  by  the  particle  after  leaving  the  plane. 

Ans.  The  directrix  intersects  the  plane  at  the  point  where  the  particle 
began  its  motion. 

(43)  A  particle  is  projected  up  a  rough  plane,  inclined  to  the 
horizon  at  an  angle  of  60',  ivith  the  velocity  ivhich  it  would  have 
acquired  in  falling  freely  through  a  space  of  12  ft.,  and  just  reaches 
the  top  of  the  plane.  Find  the  altitude  of  the  plane,  its  roughness 
being  such  that  if  it  were  inclined  to  tlie  horizontal  at  an  angle  of 
30°,  tlie  particle  would  be  on  the  point  of  sliding. 

Ans.  9  ft. 

(44)  A  ring  slides  dmvn  a  straight  rod  whilst  the  rod  is  carried 
uniformly  in  one  plane,  at  a  given  angle  to  the  horizon.  Find  the 
path  described  by  the  ring. 

Ans.  A  parabola. 

(45)  A  given  circle  and  a  given  point  are  in  the  same  vertical 
plane,  the  point  being  within  the  circle.  Find  the  straight  line  of 
quickest  descent  from  the  point  to  the  circle. 

Ans.  The  required  straight  line  is  the  distance  between  the  given  point  and 
the  lower  end  of  that  chord  of  the  circle  which  passes  through  the  given 
point  and  terminates  at  the  highest  point  of  the  circle. 

f46)  A  given  point  and  a  given  straight  line  are  in  the  same 
vertical  plane.  Determine  the  straight  line  of  quickest  descent  from 
the  given  line  to  the  point. 

Ans.  From  the  given  point  P  draw  PX  horizontally  to  meet  the  given  line 
at  X.  Draw  upwards  along  the  line  a  length  XFequal  to  PX.  The  straight 
line  joining  P  and  Y  is  the  required  straight  line. 


166  KINEMATICS   OF   A   POINT — TRANSLATION. 

(47)  To  determine  the  straight  line  of  slowest  descent  from  a  given 
point  to  a  given  circle,  the  point  being  without  the  circle  and  both  in 
the  same  vertical  plane,  the  highest  point  of  the  circle  being  lower 
than  the  given  point. 

Ans.  Let  Pbe  the  given  point  and  Q,  tlie  highest  point  of  the  circle.  Join 
PQ  and  produce  it  to  cut  the  circle  at  B.     Then  PM  is  the  line  required. 

(48)  Determine  the  straight  line  of  sloivest  descent  from,  a  given 
circle  to  a  given  point  without  it,  the  point  and  circle  being  in  tlie 
same  vertical  plane,  and  the  point  being  lower  than  tJie  lowest  point 
of  the  circle. 

Ans.  From  the  given  point  draw  an  indefinite  straight  line  cutting  the 
circle  at  its  lowest  point,  and  the  second  intersection  of  the  indefinite  line  and 
the  circle  is  the  required  line. 

(49)  Find  the  straight  line  of  slowest  descent  from  one  given 
circle  to  another,  both  circles  being  in  the  same  vertical  plane,  and 
each  exterior  to  the  other,  the  highest  point  of  the  latter  circle  being 
lower  than  the  lowest  of  the  former. 

Ans.  Produce  the  line  which  joins  the  lowest  point  of  the  first  circle  and 
•the  highest  point  of  the  second,  to  meet  both  circles  again.  The  distance  be- 
tween the  second  intersections  is  the  required  line. 

(50)  A  smooth  tube  of  uniform  bore  is  bent  into  the  form  of  a 
circular  arc  greater  than  a  semicircle,  and  placed  in  a  vertical 
plane  tvith  its  open  ends  upwards  and  in  the  same  horizontal  line. 
Find  the  velocity  with  which  a  ball  that  fits  the  tube  must  be  pro- 
jected along  the  interior  from  the  lowest  point,  in  order  that  it  may 
pass  out  at  one  end  and  re-enter  at  the  other. 

Ans.  If  r  is  the  radius,  h  the  depth  of  the  centre  of  the  circle  below  the 
horizontal  through  the  two  ends,  Vi  the  required  velocity, 

«,s  =  |(r2  +  mr  +  27t2). 

(51)  A  particle  slides  from  rest  down  a  smooth  tube  in  the  form 
of  the  thread  of  a  screw  the  axis  of  which  is  vertical.  Find  the 
time  in  which  it  voill  make  a  complete  revolution  about  the  axis. 

Ans.  If  r  is  the  radius  of  the  cylinder  on  which  the  helix  is  described,  and 
4X  the  angle  which  the  thread  makes  with  the  generating  line  of  the  cylinder, 
the  required  time  is 

/     8;rr    U 

\g  sin  'Za  j 

(52)  A  particle  falls  to  the  lowest  point  of  a  cycloid  down  any  arc 
of  the  curve,  the  axis  of  the  cycloid  being  vertical  and  its  vertex 
dowmvards.  Prove  that  the  vertical  velocity  is  greatest  when  it  has 
completed  half  its  vertical  descent. 

(53)  Also  prove  that  it  describes  half  the  path  in  two  thirds  of 
the  whole  time. 

(54)  If  a  clock  pendulum  lose  5  sec.  a  day,  determine  the  altera- 
tion tvhich  should  be  made  in  its  length. 

Ans.  It  should  be  diminished  by  nearly  the  5^-77.  part  of  its  length. 

(55)  A  seconds  pendulum  was  too  long  on  a  given  day  by  a  quan- 
tity a.    It  was  then  over-corrected  so  as  to  be  too  short  by  a  during 


MISCELLANEOUS   PROBLEMS.  167 

the  next  day.  Prove  that,  I  being  the  length  of  the  seconds  pendulum, 
the  number  of  minutes  gained  in  the  two  days  was  nearly 

1080^. 

(56)  A  seconds  pendulum  earned  to  the  top  of  a  mountain  is 
found  to  lose  there  43.2  sec.  a  day.    Find  the  height  of  the  mountain, 

supposing  the  radius  of  the  earth  to  be  4000  miles. 
Ans.  2  miles. 

(57)  Find  the  time  of  vibration  of  a  pendulum  20  feet  long. 

Ans.  Approximately  2.5  seconds. 

(58)  A  body  dropped  from  the  top  of  a  wall  falls  to  the  ground 
while  a  pendulum  6  inches  long  makes  5  beats.  Find  the  height  of 
the  wall. 

Ans.  -j-Tc'^  feet. 
4 

(59)  A  seconds  pendulum  is  lengthened  one  hundredth  of  an  inch. 
Find  how  many  seconds  it  will  lose  daily. 

Ans.  About  11  seconds. 

(60)  J^  the  length  of  a  seconds  pendulum  is  39.1386  inches,  what 
will  be  the  length  of  one  which  vibrates  40  times  a  minute. 

Ans.  88.06185  inches. 

(61)  If  the  length  of  a  seconds  pendulum  is  39.1393  inches,  find 

the  valu£  of  g. 

Ans.  g  —  32.190  feet-per-sec,  per  sec. 

(62)  A  pendulum  which  beats  seconds  at  the  equator  gains  5 
mimdes  a  day  at  the  pole.    Compare  polar  and  equatorial  gravity. 

Ans.  144  to  145,  approximately. 

(63)  Two  pendulums  the  lengths  of  ivhich  are  U  and  h  vibrate  at 
different  points  on  the  earth's  surface.  The  number  of  vibrations 
which  they  make  in  the  same  time  are  in  the  ratio  mi  to  m,.  Find 
the  ratio  of  g  at  the  tivo  places. 

Ans. J. 

(64)  A  seconds  pendulum  is  carried  to  the  top  of  a  mountain  of 
tchich  the  height  w  1  mile.  Find  the  number  of  seconds  it  unit  lose 
daily,  gravity  being  supposed  to  vary  inversely  as  the  square  of  the 
distance  from  the  centre  of  the  earth,  and  the  radius  of  the  earth  to 
be  4000  miles. 

Ans.  About  21.6  seconds. 

(65)  A  body  revolves  uniformly  in  a  circle  of  4  ft.  radius  in  10 
seconds.  Find  the  angular  velocity  and  the  velocity  at  the  circum- 
ference. 

Ans.  0.628  radian  per  sec;  2.5  ft.  per  sec. 

(66)  A  body  revolves  uniformly  in  a  circle  of  4  ft.  radium  with  a 
velocity  of  8  ft.  per  sec.     Find  the  normal  acceleration. 

Ans.  16  ft.-per-sec.  per  sec. 

(67)  A  body  revolves  imiformly  in  a  circle  of  18  ft.  radius.  The 
normal  acceleration  is  5  ft.-per-sec.  per  sec.    Find  the  velocity. 

Ans.  9.5  ft.  per  sec. 


168  KINEMATICS  OF  A   POINT — TRANSLATION. 

(68)  A  body  whose  velocity  is  10  ft.  per  sec.  is  made  to  move  uni- 
formly in  a  circle  by  a  normal  acceleration  of  2  ft.-per-sec.  per  sec. 
Find  the  radius. 

Ans.  50  ft. 

(69)  Two  points  have  velocities  Vi  and  «»  and  are  made  to  move  in 
a  circle  by  reason  of  central  accelerations  inversely  jproportional  to 
the  square  of  the  distance  from  the  centre.  The  distance  of  one 
point  is  ri.    Find  the  distance  ra  of  the  other. 

Ans.  —  :  —  =  ra*  :  1 
Tt      ra 


(70)  Find  the  relation  between  the  distances  r\  and  ra  and  the 
times  of  revolution  ti  and  ta. 


=  r,« :  r,«;    «,* :  <,«  =  n^ :  r«» 

n  Ta 


1 


I 


KINEMATICS  OF  A  EIGID  SYSTEM. 


CHAPTEK  I. 

RIGID  SYSTEM  WITH  ONE  POINT  FIXED. 


ROTATION.  ANGULAR  DISPLACEMENT.  LINEAR  DISPLACEMENT  IN  TERMS  OP 
ANGULAR.  LINE  REPRESENTATIVE  OP  ANGULAR  DISPLACEMENT.  RESO- 
LUTION AND  COMPOSITION  OP  ANGULAR  DISPLACEMENTS.  ANGULAR 
VELOCITY.  INSTANTANEOUS  AXIS  OP  ROTATION.  ANGULAR  ACCELERA- 
TION. RESOLUTION  AND  COMPOSITION  OP  ANGULAR  VELOCITIES  AND 
ACCELERATIONS.  EQUATIONS  OP  MOTION  OP  A  POINT  OP  A  ROTATING 
SYSTEM.  MOMENT  OP  ANGULAR  VELOCITY  AND  ACCELERATION.  GENERAL 
ANALYTICAL  DETERMINATION  OP  RESULTANT  FOR  CONCURRING  ANGULAR 
VELOCITIES  AND  ACCELERATIONS. 

Rotation. — When  a  rigid  system  moves  so  that  all  its  points 
describe  circles  in  parallel  planes  about  a  common  straight  line  or 
axis  passing  through  the  centres  of  the  circles  and  perpendicular  to 
their  planes,  the  system  is  said  to  rotate  or  have  a  motion  of  rotation 
about  that  axis.  Any  plane  parallel  to  the  planes  of  the  circles  is 
the  plane  of  rotation. 

Since  the  system  is  rigid,  every  point  must  describe  its  circle  in 
the  same  time,  or  the  angular  speed  (page  72)  of  every  point  is  the 
same. 

If  the  angular  speed  does  not  change  and  the  plane  of  rotation 
does  not  change,  the  rotation  of  the  system  is  uuiform.  If  either 
the  angular  speed  changes  or  the  plane  of  rotation  changes,  the 
rotation  is  variable. 

Motion  of  a  Rigid  System  with  One  Point  Fixed. — We  have  de- 
fined translation  (page  13)  as  motion  of  a  system  such  that  every 
straight  line  joining  every  two  points  remains  always  parallel  to 
itself  during  the  motion.  In  such  case  the  motion  of  the  system  is 
the  same  as  that  of  any  one  of  its  points,  and  the  study  of  the  trans- 

*  The  advanced  student  sliould  read  this  portion  of  the  worl<  in  connection 
with  the  analogous  portions  of  Statics  referred  to  in  the  text.  The  student  new 
to  the  subject  would  do  well  to  omit  this  portion  of  the  work  and  take  it  in 
connection  with  Statics  later. 

169 


170        RIGID  SYSTEM  WITH  OXK  I'OINT  FIXED— ROTATION.    [CHAP.  I. 

lation  of  a  system  is  the  same  as  the  study  of  the  translation  of  a 
point.  In  the  preceding  Chapters  we  have  treated  of  the  kinematics 
of  a  point  or  translation. 

If  one  point  of  a  system  is  fixed,  the  motion  of  which  it  is  capa- 
ble will  be  more  or  less  complex  according  as  its  points  can  or  can 
not  move  relatively  to  each  other.  We  restrict  our  discussion  to 
rigid  systems,  that  is,  systems  whose  points  can  not  move  relatively 
to  each  other. 

If  one  point  of  such  a  system  is  fixed,  there  can  be  no  translation 
and  the  only  motion  of  which  it  is  capable  is  one  of  rotation  as  just 
defined. 

In  such  motion  it  is  evident  that  all  straight  lines  in  the  system 
must  remain  straight  lines  of  unchanged  length  and  mutual  inclina- 
tion, and  all  planes  must  remain  planes  of  unchanged  form,  area 
and  mutual  inclination.  Also  the  motions  of  any  two  points  indefi- 
nitely near  must  be  indefinitely  nearly  the  same. 

Angular  Displacement  of  a  Rigid  System. — Let  AB  be  the  axis 
of  rotation  of  a  rigid  system.  Then,  since  every  point  must  com- 
plete its  circle  in  the  same  time,  the  angle  de- 
scribed by  all  points  in  any  given  time  must 
be  the  same  as  the  angle  described  by  any 
one  point  P. 

The  angle  Q  between  the  initial  and  final 
positions,  in  any  given  time,  of  the  perpendic- 
ular PO  from  any  point  Pto  the  axis,  is  called 
the  angular  displacement  of  the  system. 

Since  the  angle  6  is  measured  in  radians,  it  is  independent  of 
the  length  PO  (page  5). 

Linear  Displacement  in  Terms  of  Angular.— Let  OPi  =  OPi  =  r 
be  the  radius  for  any  point  P  which  moves  in  a  circle  perpendicular 
to  the  axis  at  O,  through  the  angular  displace- 
ment PiOPa  =  <9,  from  the  initial  position  Pi 
to  the  final  position  Pa.  Then  the  triangle 
PiOPi  is  isosceles,  and  if  we  draw  ON  per- 
pendicular to  P1P2  we  have  the  linear  dis- 
placement 

PiP3=2r  sin  4. 

O 

Line  Representative  of  Angular  Displacement  of  a  Rigid  System. 
— An  angular  displacement  of  a  rigid  system  is  given  when  we 
know  not  only  its  magnitude  and  the  direction  of  rotation  in  the 
plane  of  rotation,  but  also  the  direction  of  that  plane  in  space. 

It  is  therefore  a  vector  quantity  having  not  only  magnitude  and 

sign,  but  also  direction,  and  it  can 
be  completely  represented  by  a 
straight  line,  like  linear  displace- 

^        ment  (page  34). 

Thus  the  length  of  the  straight 
line  AB  denotes  the  magnitude  of 
the  angular  displacement  P1OP1 
=  0.  The  plane  of  rotation  is  at 
right  angles  to  the  line  AB,  ivhich  is  therefore  coincident  tvith  the 
axis  of  rotation.  The  direction  of  rotation  is  always  clockwise 
when  we  look  along  this  line  in  the  direction  indicated  by  the 
arrow. 

By  direction  of  an  angular  displacement  we  mean  always  the 
direction  of  its  line  representative  as  denoted  by  the  arrow. 


CHAP.  I.]  ANGULAR    DISPLACEMENTS.  171 

Composition  and  Resolution  of  Successive  Angular  Displacements. 
— Let  a  rigid  system  with  one  point  fixed  undergo  successive  angular 
displacements.  It  is  required  to  determine  the  resultant  angular 
displacement. 

Evidently  the  successive  angular  displacements  may  be  about 
the  same  or  about  different  axes,  and  in  either  case  may  be  finite  or 
indefinitely  small. 

(a)  About  the  Same  Axis — Finite  or  Indefinitely  Small. — If  the  axis 
of  all  the  angular  displacements  is  the  same,  the  plane  of  rotation 
does  not  change,  and  the  magnitude  and  sign  or  direction  of  the 
resultant  displacement  in  that  plane  is  given  by  the  algebraic  sum 
of  the  magnitudes  of  the  successive  displacements,  whether  they 
are  finite  or  indefinitely  small 

Inversely,  any  angular  displacement  about  a  given  axis  may  be 
resolved  into  any  number  of  successive  displacements  about  the 
same  axis,  whether  finite  or  indefinitely  small,  provided  the  alge- 
braic sum  of  their  magnitudes  is  equal  to  the  magnitude  of  the  given 
displacement  and  the  same  in  sign. 

(b)  About  Different  Axes— Displacements  Finite. — The  axes  must 
pass  through  the  fixed  point  of  the  system.  First  let  the  displace- 
ments be  finite. 

Let  O  be  the  fixed  point  of  the  system,  and  ORi ,  ORi  the  initial 
positions  of  the  two  given  axes.  Take  ORi  —  ORi,  and  let  us  sup- 
pose first  a  displacement  d,  of  the  system  about  ORi  and  then  a  dis- 
placement fj2  about  the  new  position  of  the  other  axis.  During  this 
motion  R2  and  Ri  will  move  on  the  surface  of  a  sphere. 

When  the  system  is  rotated  an  angle  9i  about  ORi,  the  axis  ORt 
moves  from  ORi  to  OR^'. 

Now  join  R1R2,  R1R2'  and  RiR^  by 
great  circles  of  the  sphere.  Then  the 
angle  R^RiRi'  =  Oi. 

Bisect  this  angle  by  a  great  circle 
meeting  Ri'R-.  at  D.  Draw  a  great 
circle  through  R^'  inclined  to  Ri'Ri  at 

the  angle  ^  and  meeting  RiD  at  R. 

Then  draw  Ri'Ci,  making  the  same 
angle  ^  with  R^'Rx  on  the  other  side, 
and  make  R^Ci  =  R^R.    Then  i2,a  will  equal  RxR ,  and  the  angle 
Ri'RiC-i  =  ^■ 

When  then  the  system  is  rotated  about  ORi ,  and  the  axis  OR^ 
moves  to  OR/  through  the  angle  R^RM/  =  fii,  the  Ime  OR  will 
move  to  OC2  through  the  angle  RRxd  =  0i. 

If  now  the  system  is  rotated  about  OR2  through  the  angle 
CiR'R  =  0-2,  the  line  OC2  moves  back  to  OR. 

Hence  tho  line  OR  has  the  same  position  before  and  after  the 
rotations.    The  resultant  displacement  is  then  a  displacement  about 

Hence,  the  resultant  of  two  successive  rotations  6,  about  ORi  and 
«2  about  OR2 ,  when  the  axes  intersect  in  anoint  O,  is  a  single  rota- 
tion e  about  the  axis  OR  passing  through  0. 

In  order  to  find  the  position  of  this  axis  OR  and  the  ma^itude 
of  0,  we  have  in  the  spherical  triangle  RiRRi  the  angle  RKxKi  - 

-Oi,   the  angle  RR^R  =  ^6,  and  the  exterior  angle  DRR^  =  :^^- 
Hence 


173        EIGID  SYSTEM  WITH  ONE  POINT  FIXED — ROTATION.    [CHAP.  I^ 

COS  -6  =  COS  -9,  COS  -02  —  sin  -6,  sin  -62  cos  RiBi ;     .    (1> 


sin  RiR     sin  RRi      sin  RiRi 


sin  -Q2       sin  -61         sin  -6 
2  2  2 


(2) 


Since  OJRi  and  OR2  are  lines  of  a  rigid  body  and  ORi  coincides 
with  the  position  of  the  first  axis  of  rotation  in  space,  the  second 
axis  of  rotation  in  space  has  the  position  OR-i'  and  not  OR2*  Hence, 
in  general,  the  order  of  the  two  successive  rotations  is  not  indifferent. 

Example. — The  telescope  of  a  theodolite,  originally  horizontal 
and  pointing  north,  is  first  turned  into  an  altitude  of  60°  and  then 
turned  towards  the  west  into  the  prime  vertical.  Find  the  resultant 
rotation. 

Ans.  We  have  Qj  =  60°,     62  =  90°,     BiB^  =  90°.     Hence 


cos  1 0  =  y  8  X  |/|  =  i/|,     or    Sin  i»  =  i/|. 
For  the  position  of  the  axis  we  have 

itB= ^=24/-^,     sini2i?2  = ^  =  4/%. 

2^2  2^2 


sini2i 


If  we  invert  the  order  of  the  two  rotations,  we  have  9i  =  90°,  62  =  60" 
B1B2  =  90°.     Hence 

cos  -0  =  g-y  ^'      or    sin  -9  =  2^2'  ^  ^^^^^^   ' 
For  the  position  of  the  axis 

sin  BiB  =  — —  =  i/i,    sin  BB^  = —  =  2  a/I. 

2  r  ^  2  r  2 

(c)  About  Different  Axes — The  Displacements  Indefinitely  Small. — 

A ^— ^^c    ^®cond   let   the   rotations  be  indefi- 

K/^  ^„.,-''^/         nitely  small.     Let   OA  =  9   and   OB 

y^      ^      ^X  =  0  be  the  line  representatives.  Com - 

/y^-^"^^"^  '''  plete    the    parallelogram    and    draw 

-,^d_5_ ^'''  ^G-      Let  P  be  any  point  of  the  sys- 

"       s  B  tern  in  the  plane  of  OA  and  05,  and 

draw  the  perpendiculars  P§,  PR,  PS.  When  rotation  occurs- 
about  OA,  the  point  P  will  move  perpendicularly  to  the  plane  of  the 
paper  through  a  very  small  distance  represented  by  rO  or  OA  x  PQ' 
(page  5). 

When  rotation  occurs  about  OB,  the  point  P  moves  perpendic- 
ularly to  the  plane  of  the  paper  also,  through  a  very  small  distance 
represented  by  OB  x  PS.  Since  both  these  displacements  are  very 
small  they  coincide  in  direction,  and  the  resultant  is 

OAx  PQ  +  OB  X  PS=OC  X  PR. 

Hence  the  resultant  displacement  is  given  by  OC. 


•€HAP.   I.]  AXIS   OF   ROTATION.  173 

We  should  have  the  same  result  if  the  rotation  about  OB  occurred 
first,  also  if  the  point  P  had  been  taken  within  the  angle  AOB. 
Also  whether  OA  and  OB  are  axes  fixed  in  the  body  or  in  space. 

If  we  have  more  than  two  successive  rotations,  the  third  may  be 
compounded  with  the  resultant  of  the  first  two  in  like  manner. 

Hence  if  a  rigid  system  with  one  point  fixed  undergo  any  num- 
ber of  successive  indefinitely  small  angular  displacements  about 
different  axes  either  fixed  in  the  system  or  fixed  in  space,  the 
resultant  angular  displacement  is  obtained  by  treating  the  line 
representatives  precisely  like  linear  displacements  (page  35).  We 
have  thus  the  parallelogram  and  polygon  of  angular  displace- 
ments. 

Composition  and  Resolution  of  Simultaneous  Angular  Displace- 
ments. — The  simultaneous  angular  displacements  may  be  finite  or 
indefinitely  small  and  must  be  either  about  the  same  axis  or  differ- 
ent axes. 

(a)  About  the  Same  Axis.— If  the  axis  of  all  the  angular  displace- 
ments is  the  same,  the  plane  of  rotation  does  not  change  and  the 
magnitude  and  sign  or  direction  of  the  resultant  displacement  in 
that  plane  is  given  by  the  algebraic  sum  of  the  magnitudes  of  the 
simultaneous  angular  displacements,  whether  they  are  finite  or  in- 
definitely small. 

(b)  About  Diflferent  Axes. — If  the  rotations  OA,  OB  are  indefi- 
nitely small,  we  see  from  the  figure,  page  171,  that  it  makes  no  differ- 
ence whether  they  are  successive  or  simultaneous.  We  can  resolve 
and  combine  them,  therefore,  by  their  line  representatives  just  like 
linear  displacements  (page  35).  We  have  then  the  parallelogram 
and  polygon  of  angular  displacements. 

If  the  rotations  OA,  OB  are  finite,  we  can  divide  each  up  into  a 
number  of  indefinitely  small  rotations  and  treat  each  pair  as  before. 
We  have  then  the  parallelogram  and  polygon  of  angular  displace- 
ments in  this  case  also. 

Composition  and  Resolution  of  Angular  Displacements  in  gen- 
eral.— We  see  then  that  in  all  cases  except  finite  successive  angular 
displacements  about  different  axes  we  can  combine  and  resolve 
any  number  of  angular  displacements  whether  simultaneous  or 
successive,  finite  or  indefinitely  small,  about  the  same  or  about 
different  axes  by  means  of  the  line  representatives,  just  like  linear 
displacements. 

Simultaneous  angular  displacements  are  usually  called  com- 
ponent angular  displacements.  Component  angular  displacements 
must  then  be  understood  to  always  mean  simultaneous  angular  dis- 
placements, unless  otherwise  specified. 

Sign  of  Components  of  Angular  Displacement. — The  sign  of  the 
line  representatives  of  the  components  along  the  axes  X,  Y,  Z  of  an 
angular  displacement  follows  the  same  rule  as  for  linear  displace- 
ment fpage  36).  Hence  if  we  look  along  the  line  representatives 
towards  the  origin,  the  radius  vector  wUl  always  be  seen  to  move 
connter-clochwise. 

Axis  of  Rotation.— In  every  possible  dis- 
placement of  a  rigid  system  with  one  point 
fixed,  there  is  one  line  fixed  in  the  system  pass^ 
ing  through  the  fixed  point,  called  the  axis  <^ 
rotation,  which  has  the  same  position  in  both 
the  initial  and  final  positions  of  the  system. 

Let  O  be. the  fixed  point  of  the  system,  and     .  .        ._ 
let  A,,  Bi  be  the  initial  and  Aa,  B^  the  final  positions  m  space  of 
two  points  of  the  system. 


174        RIGID  SYSTEM  WITH  ONE  POINT  FIXED — ROTATION.    [CHAP.  I^ 

Since  the  system  is  rigid,  OAi  =  OA^ ,  and  OBi  —  OB 2.  Let  Ai 
be  brought  to  Ai  by  rotation  about  an  axis  through  O  perpendicu- 
lar to  the  plane  of  AiOAt.  By  this  rotation  Bi  moves  to  6,  and  since 
the  system  is  rigid,  Oh  =  OBi  —OB^,  and  A^b  —  A2B2.  The  triangles 
OAib  and  OA-iB^  are  then  equal  in  all  respects,  and  6  can  be 
brought  to  JSa  by  rotation  about  OA-i.  The  given  displacement  can 
always  then  be  produced  by  two  successive  rotations  about  two 
axes  passing  through  O.  As  we  have  seen  (page  171),  two  such 
successive  rotations  give  as  resultant  a  single  rotation  about  an 
axis  through  O.     This  is  the  axis  of  rotation. 

Cor.  1.  Hence  any  angular  displacement  of  a  rigid  system  with 
one  point  fixed  is  completely  specified  by  the  line  representative  of 
the  resultant  angular  displacement,  which  coincides  in  direction 
with  the  axis  of  rotation. 

CoR.  3.  Any  angular  displacement  of  a  rigid  system  with  one 
point  fixed  may  be  resolved  into  three  angular  displacements  about 
the  co-ordinate  axes  through  the  fixed  point  taken  as  origin. 

Cor,  3.  Every  line  in  the  system  parallel  to  the  axis  of  rotation 
remains  unchanged  in  direction. 

Mean  Angular  Velocity  of  a  Rigid  System. — The  magnitude  of 
the  angular  displacement  during  a  given  time  of  a  rigid  system 
with  one  point  fixed,  divided  by  the  number  of  units  of  time,  gives 
the  magnitude  of  the  mean  angular  velocity  of  the  system. 

It  is  represented  by  a  line  just  like  angular  displacement  (page 
170).  By  direction  of  mean  angular  velocity  we  always  mean  direc- 
tion of  the  line  representative. 

Mean  angular  speed  then  is  mean  time-rate  of  angle  described 
(page  72).  Mean  angular  velocity  is  mean  time-rate  of  angular 
displacement. 

Instantaneous  Angular  Velocity  of  a  Rigid  System. — The  limit- 
ing magnitude  and  direction  of  the  mean  angular  velocity  when 
the  interval  of  time  is  indefinitely  small  is  the  instantaneous  angu- 
lar velocity. 

The  term  angular  velocity  always  signifies  instantaneous  angu- 
lar velocity  unless  otherwise  specified. 

It  may  be  represented  by  a  straight  line  just  like  angular  dis- 
placement (page  170;. 

By  direction  of  an  angular  velocity  we  always  mean  the  direc- 
tion of  its  line  representative. 

We  see  then  that  angular  displacement  and  angular  velocity 
are  vector  quantities  like  linear  displacement  and  linear  velocity. 
Angular  velocity  is  directed  angular  speed,  just  as  linear  velocity 
is  directed  linear  speed.  Speed  is  magnitude  of  velocity,  whether 
linear  or  angular  (page  43). 

Instantaneous  Axis  of  Rotation.  —  The  instantaneous  angular 
velocity  of  a  rigid  system  is  then  given  by  its  line  representative. 
This  line  representative  coincides  in  position  with  the  axis  of  rota- 
tion at  the  instant.  This  axis  is  then  the  instantaneous  axis  of  rota- 
tion. 

Unit  of  Angular  Velocity. — Since  the  magnitude  of  the  angular 
velocity  at  any  instant  is  the  angular  speed  in  a  given  direction  at 
that  instant,  the  unit  of  angular  velocity  is  the  same  as  for  angul;  r 
speed,  or  one  radian  per  sec.  We  denote  the  magnitude  then  by 
the  same  letter,  oa,  and  we  have  the  same  numeric  equations  as  for 
angular  speed  (page  73). 

Thus  for  mean  angular  velocity 

CO    =  T 1 W 


CHAP.  I.]  ANGULAR  ACCELERATION.  175 

and  for  instantaneous  angular  velocity 

-  =  * («) 

TTnifonn  and  Variable  Angular  Velocity. — Angular  velocity  is 
uniform  when  the  line  representative  has  the  same  magnitude  and 
direction  whatever  the  interval  of  time.  Uniform  angular  velocity 
is  then  uniform  angular  speed  in  an  unchanging  plane,  just  as  uni- 
form linear  velocity  is  uniform  linear  speed  in  an  unchanging 
direction  (page  43). 

In  such  case  angular  velocity  is  the  same  as  the  mean  angular 
velocity  for  any  interval  of  time. 

When  eivher  the  magnitude  or  direction  of  the  angular  velocity 
changes  it  is  variable. 

W  hen  the  magnitude  alone  changes  we  have  variable  angular 
speed  in  an  unchanged  plane  of  rotation.  When  the  direction  only 
changes  we  have  uniform  angular  speed  in  a  changing  plane  of 
rotation.  When  both  change  we  have  variable  angular  speed  in  a 
changing  plane  of  rotation. 

Mean  Angular  Acceleration  of  a  Rigid  System. — If  OA  =  co,  and 
OB  —  CO  are  the  line  representatives  of  the  initial  and  final  angular 
velocities  of  a  rigid  system  with  one  point  fixed, 
during  any  time  t,  then  AB  is  the  line  represent- 
ative of  the  integral  angular  acceleration  of  the 

AB 
system  during  the  time  t,  and  —r-  gives  the 

magnitude  of  the  mean  angular  acceleration  whose  direction  is  AB. 
(Compare  page  48.) 

Mean  angular  acceleration  then  is  time-rateof  change  of  angular 
velocity,  whether  that  change  takes  place  in  the  direction  of  tJie  an- 
gular velocity  or  not. 

Instantaneous  Angular  Acceleration  of  a  Rigid  System. — The 
limiting  magnitude  and  direction  of  the  mean  angular  acceleration 
when  the  interval  of  time  is  indefinitely  small  is  the  instantaneous 
angular  acceleration.  It  is  the  limiting  time-rate  of  change  of  an- 
gular velocity  whether  that  change  takes  place  in  the  direction  of 
the  angular  velocity  or  not. 

Angular  acceleration  always  signifies  instantaneous  angular 
acceleration  unless  otherwise  specified.  ,      j. 

It  may  be  represented  by  a  straight  line  just  hke  angular  dis- 
placement (page  170).  By  direction  of  an  angular  acceleration  we 
mean  the  direction  of  its  line  representative. 

Instantaneous  Axis  of  Angular  Acceleration.— The  instantaneous 
angular  acceleration  of  a  rigid  system  is  then  given  by  its  line  rep- 
resentative. This  line  representative  coincides  m  position  witn 
the  axis  of  angular  acceleration  at  the  instant.  This  axis  is  then 
the  instantaneous  axis  of  angular  acceleration. 

Angular  acceleration  may  be  zero,  uniform  or  variable.  VV  hen 
it  is  zero,  the  angular  velocity  is  uniform  and  we  have  uniform 
angular  speed  and  an  unchanging  plane  of  rotation. 

When  It  is  uniform,  it  has  the  same  magnitude  and  the  same 
direction  whatever  the  interval  of  time.  In  such  case  the  accelera- 
tion is  equal  to  the  mean  acceleration  for  a^J  .^"t^rval  of  time  It 
the  direction  coincides  with  that  of  the  i^^t^^l  velocity,  we  have 
uniform  rate  of  change  of  angular  speed  and  ^n  unchanged  plane 
of  ro:ation.  If  it  makes  an  angle  with  the  velocity,  we  have  a 
changing  plane  of  rotation  and  variable  velocity. 


176       RIGID  SYSTEM  WITH  ONE  POINT  FIXED — ROTATION.    [CHAP.  I. 

When  it  is  variable,  either  direction  or  magnitude  changes  or 
Tjoth  change. 

If  the  angular  acceleration  is  always  at  right  angles  to  the  an- 
gular velocity,  it  only  changes  the  direction  but  not  the  magnitude 
of  that  velocity. 

Hence,  just  as  on  page  53  a  normal  linear  acceleration  has  no 
effect  upon  the  linear  speed,  but  only  changes  the  direction  of  mo- 
tion, so,  if  a  rigid  system  rotating  with  given  angular  speed  about 
an  axis  has  an  angular  acceleration  about  an  axis  always  perpendic- 
ular to  the  first,  there  is  change  of  direction  of  this  axis  but  no 
change  of  angular  speed  about  it. 

The  gyroscope  is  an  illustration  of  this  principle. 

Resolution  and  Composition  of  Angular  Velocity  and  Accelera- 
tion.— Since  for  an  indefinitely  small  time  the  angular  displace- 
ment is  indefinitely  small,  we  see  from  page  171  that  we  can  com- 
bine angular  velocities  and  accelerations,  whether  simultaneous  or 
successive,  by  means  of  their  line  representatives  just  like  linear 
velocities  and  accelerations  (page  43 j. 

Sign  of  Components  of  Angular  Velocity  and  Acceleration. — The 
sign  of  the  line  representatives  of  the  components  along  the  axes 
X,  Y,  Z  of  an  angular  velocity  or  acceleration  follows  the  same 
rule  as  for  linear  velocities  and  accelerations  (pages  44,  50). 

Unit  of  Angular  Acceleration. — Angular  acceleration  is  meas- 
ured in  terms  of  the  same  unit  as  rate  of  change  of  angular  speed 
(page  73),  or  one  radian-per-sec.  per  sec.  We  denote  its  magni- 
tude then  by  the  same  letter,  ex. 

Relations  between  Angular  and  Linear  Velocity  and  Accelera- 
tion.— We  have  also  the  same  relations  between  angular  and  linear 
acceleration  and  velocity  as  for  a  point  moving  in  a  circle  (page 
76). 

Thus  we  have,  for  any  point  of  a  rigid  system  whose  distance 
from  the  axis  of  rotation  at  any  distance  is  r, 

roo^=v,     ra=ft,     voo  =fn  =  roo^  =  —,    fp=ftr  =  r^a,     vr  =  r'ao. 

Equations  of  Motion  of  a  Rotating  Rigid  System  under  Diflferent 
Angular  Accelerations. — Since  angular  velocities  and  accelerations 
are  represented  by  straight  lines,  just  like  linear  velocities  and  ac- 
celerations, we  have  the  same  equations  for  motion  of  a  rotating 
rigid  system  as  on  page  50.  We  have  only  to  substitute  go  for  v,  Q 
for  s,  a  for  /. 

With  these  substitutions  equations  (1)  to  (14),  page  50,  hold  good 
and  it  is  unnecessary  to  repeat  them  here. 

Moment  of  Angular  Displacement. — Just  as  we  called  the  prod- 
uct of  the  magnitude  of  a  linear  displacement  by  the  magnitude 
of  the  perpendicular  let  fall  from  any  given  point  upon  its  du-ec- 
tion  the  moment  of  the  linear  displacement  (page  60),  so  for 
angular  displacement  we  call  the  product  of  its  magnitude  by 
the  magnitude  of  the  perpendicular  from  any  point  upon  the 
direction  of  the  line  representative  the  moment  of  the  angular 
displacement 

We  take  its  sign  just  as  for  moment  of  linear  displacement,  page 
62.  Since  the  line  representative  is  coincident  with  the  axis,  the 
perpendicular  is  the  distance  of  the  point  from  the  axis. 

Thus  if  AB  =  6  is  the  line  representative  of  an  angular  dis- 
placement OiOOi  =  0  of  a  rigid  system,  the  axis  has  the  position 


CHAP.  I.]     CONCURRIN-G  ANGULAR  DISPLACEMENTS,  ETC.  177 

AOB.  If  then  Oi  is  the  initial  position  of  any  point  of  the  sys- 
tem and  OOi  =  p  is  the  perpendic-  ^ 
ular  from  Oi  upon  the  axis  or  direc- 
tion of  the  line  representative  AB,  the 
moment  is  ±  p'j  according  to  direction, 
just  as  for  moment  of  linear  displace- 
ment (page  62;.     But  pO  is  the  length j^ ,     ^ 

of  the  arc  OiOi  described  by  the  point  '       -*         '  ^^^ 

Oi   in  a  plane   perpendicular  to   the 
plane  of  AB  and  00 1. 

Hence,  the  moment  pS  of  the  angular 
displacement  0  of  a  rigid  system  rela- 
tive to  any  point  of  the  system  gives 
the  length  of  the  arc  O1O2  described  by 
that  point  in  a  plane  perpendicular  to  the  plane  of  the  axis  AB 
and  the  radius  vector  p. 

The  corresponding  linear  displacement  of  Oi  is  evidently 

T  =  2p  sin- (1) 

Since  the  angle  OO1O2  equals  the  angle  OOaOi,  we  have  for  the 
direction  of  the  linear  displacement  relative  to  OOi , 

angleOOiO.  =  90°- 0.0r=^^ (2) 

We  have  also,  just  as  on  page  62,  the  algebraic  sum  of  the 
moments  of  any  number  of  component  angular  displacements,  rel- 
ative to  any  point,  equal  to  the  moment  of  the  resultant. 

Also,  just  as  on  page  60,  the  line  representative  of  an  angular 
displacement  may  be  laid  off  from  any  point  in  its  line  of  direction 
without  affecting  its  moment. 

Moment  of  Angular  Velocity  or  Acceleration.  —  Just  as  we 
called  the  product  of  the  magnitude  of  a  linear  velocity  or  accelera- 
tion by  the  magnitude  of  the  perpendicular  from  any  given  point 
upon  its  direction  the  moment  of  the  linear  velocity  or  acceleration, 
so  for  angular  velocity  or  acceleration  we  call  the  product  of  the 
magnitude  by  the  magnitude  of  the  perpendicular  from  any  point 
upon  the  direction  of  the  line  representative  the  moment  of  the 
angular  velocity  or  acceleration. 

We  take  its  sign  just  as  for  moment  of  linear  velocity  or  accele- 
ration (page  60).  Since  the  line  representative  is  coincident  with  the 
axis,  the  perpendicular  is  the  distance  of  the  point  from  the  axis. 

Thus  if  AB  =  (a  is  the  line  representative  of  an  angular 
velocity  of  a  rigid  system,  the  instantaneous  axis  has  the 
position  AOB.  If  then  Oi  is  any  point  of 
the  system  and  OOi  =p  is  the  perpendicu- 
lar from  Oi  on  the  axis  or  direction  of  the 
line  representative,  the  moment  is  ±  poo 
according  to  direction,  just  as  for  moment 
of  linear  velocity  (page  60).  But  this  is 
^^■^  the  linear  velocity  v  of  Oi  at  the  instant, 
in  a  direction  perpendicular  to  the  plane 
of  AB  and  00 1. 

Hence,  the  moment  poo  of  the  angidar 
velocity  co  of  a  rigid  system,  relative  to  any 
point  of  the  system  gives  the,  linear  velocity 
V  of  that  point  in  a  direction  perpendicu- 
lar to  the  plane  of  the  instantaneous  axis 
of  rotation  AB  and  the  instantaneous  radius  vector  p. 


178       BIGID  SYSTEM  WITH  ONE  POINT  FIXED — KOTATION".    [CHAP.  I^ 

In  the  same  way,  the  moment  pa  of  the  angular  acceleration  a  of 
a  rigid  system  relative  to  any  point  of  the  system  gives  the  linear- 
tangential  acceleration  ft  of  that  point  in  a  direction  perpendicular- 
to  the  plane  of  the  instantaneous  axis  of  angular  acceleration  and 
the  instantaneous  radius  vector. 

We  have  also,  just  as  on  page  62,  the  algebraic  sum  of  the  mo- 
ments of  any  number  of  component  angular  velocities  or  accelera- 
tions, relative  to  any  point,  equal  to  the  moment  of  the  resultant. 

Also,  just  as  on  page  60,  the  line  representative  of  an  angular 
velocity  or  acceleration  may  be  laid  off  from  any  point  in  its  line 
of  direction  without  affecting  its  moment. 

Concurring  Angular  Displacements,  Velocities  or  Accelerations.  — 
"We  see  then  that  angular  displacements,  velocities  or  accelera- 
tions are  represented  by  straight  lines,  called  line  representatives^ 
which  coincide  with  the  axis  of  rotation.  We  deal  with  them 
entirely  by  means  of  these  line  representatives.  When  we  speak 
of  their  "direction,"  we  mean  the  direction  of  the  line  representa- 
tives. We  resolve  and  combine  them  by  means  of  their  line  repre- 
sentatives, and  in  the  same  way  we  have  their  moments  just  as  for 
linear  displacements,  velocities  or  accelerations.  Following  the 
same  analogy,  we  can  speak  of  them  as  "applied  "  or  "acting"  at. 
certain  points.  When  they  all  intersect  at  the  same  point,  we  may 
call  them  concurring,  just  as  if  they  were  linear.  When  they  do 
not  intersect  at  the  same  point  they  are  non-concurring.  When  they 
act  in  the  same  direction  in  the  same  line  they  are  conspiring. 
When  in  the  same  or  opposite  directions  in  parallel  lines  they  are 
parallel.  When  in  opposite  directions  in  the  same  line  or  in  paral- 
lel lines  they  are  opposite.  When  they  lie  in  the  same  plane  they 
are  co-planar. 

Condition  for  Rotation  only.  —If  a  rigid  system  has  one  point 
fixed,  it  can  have  no  translation  but  only  rotation,  and  therefore 
all  the  component  angular  displacements,  velocities  or  accelerations 
must  reduce  to  a  concurring  system,  so  that  we  have  a  single  result- 
ant angular  displacement,  velocity  or  acceleration  about  an  axis 
through  this  point,  which  is  therefore  at  rest. 

General  Analytical  Determination  of  Resultant  Angular  Dis- 
placement, Velocity  or  Acceleration  for  any  Number  of  Concurring 
Components. — We  see  then  that  all  the  equations  of  pages  63  to  65 
hold  good  for  angular  displacements,  velocities  or  accelerations,  as 
well  as  for  linear. 

For  angular  displacements  we  have  only  to  substitute  6  in  place 
of  V.  The  moments  Mx,  My,  Mz  then  give  the  arcs  of  displacement 
about  the  axes  of  X,  F,  Z  of  the  origin,  considered  as  a  point  of 
the  rigid  system,  rotating  about  the  resultant  axis. 

For  angular  velocities  we  have  only  to  substitute  oo  for  v.  The 
moments  Mx ,  My,  Mz  then  give  the  component  linear  velocities  Vx , 
Vy ,  Vz  along  the  axes  of  X,  F,  Z  of  the  origin,  considered  as  a  point 
of  the  rigid  system,  rotating  about  the  resultant  axis. 

For  angular  accelerations  we  have  only  to  substitute  a  for  u. 
The  moments  Mx,  My,  Mz  then  give  the  component  linear  tangential 
accelerations  ftx,  fty,  ftz  along  the  axes  of  X,  Y,  Z  of  the  origin, 
considered  as  a  point  of  the  rigid  system,  rotating  about  the  re- 
sultant axis. 

To  make  our  notation  consistent  we  should  also  replace  cos  a,, 
cos  h,  cos  c,  page  65,  by  cos  d,  cos  e,  cos  /,  and  replace  cos  d,  cos  e» 
cos  f,  page  66,  by  cos  a,  cos  6,  cos  c. 

We  have  then  from  page  65,  equation  (4),  for  the  component. 
linear  velocities  Vx,Vy,  Vz  along  the  axes  of  X,  Y,  Z  of  the  origiUy 


CHAP.  I.]         TWO   CONCUERING  ANGULAB  VELOCITIES. 


179 


considered  as  a  point  of  the  rigid  system,  rotating  about  the  re- 
sultant axis, 


Vx  =  oozy  —  aoyZ ; 

Vy  =  OOxZ  —  00zX\ 
Vz    =    Oi>yX  —    COxy- 

We  have  also  in  the  same  way 

fix  =  azy  —  ayZ  ; 
fty  =  axZ  —  azX  ; 
ftz  =  cXyX  —  axy. 


(1) 


(2) 


Equations  (2)  give  the  component  linear  tangential  accelerations 
along  the  axes  of  X,  Y,  Z  of  the  origin,  considered  as  a  point  of  the 
rigid  system,  rotating  about  the  resultant  axis. 

If  we  multiply  the  first  of  equations  (1)  by  eox ,  the  second  by 
coy ,  the  third  by  a>z  and  add,  we  obtain 


VxOOx    +    VyOOy    +    VzOOz    =  0. 


(3) 


Equation  (3)  is  the  condition  for  rotation  only.  When  it  is  ful- 
filled, we  know  that  the  motion  of  the  system  is  that  of  rotation 
only  about  the  instantaneous  axis. 

Resultant  of  Two  Concurring  Component  Angular  Displacements, 
Velocities  or  Accelerations.*— It  will 
be  of  profit  to  specially  discuss  the 
case  of  two  concurring  component 
angular  displacements,  velocities  or 
accelerations. 

Let  the  two  angular  velocities  oji, 
GOi  be  in  the  same  plane  and  pass 
through  the  points  A  and  B  of  a 
rigid  system,  so  that  they  intersect 
at  O.  Then  the  resultant  oo,-  must 
pass  through  O  and  be  in  the  plane 

of  CiJi,  ooi. 

Take  any  point  P  in  this  plane 
and  draw  the  perpendiculars  Pwi  = 
p,,  Pn.=pi,  Pn=pr.  Then,  since 
the  moment  of  the  resultant  is  equal 
to  the  algebraic  sum  of  the  moments 
of  the  components, 

GOrPr  =  0Oipi    +   OOiPi,  .       (1) 

where  regard  must  be  paid  to  the 
signs  in  any  case.  Thus  we  have  in 
the  figure 

00,p,-  =  QOipi  —  OO^Pt. 

Draw  the  line  AB,  intersecting  the 
resultant  oor  at  the  point  C  Let  <:i-i 
be  the  angle  of  «,,  and  a^  the  angle  of 
<a,  with  AB.  If  we  take  moments 
about  C,  we  have 

oo^ .  ACsinai  =  oo^  .EC  sin  a-,. 


*  Compare  Statics— Non-concurring  Forces. 


180        KIGID  SYSTEM  WITH  ONE  POINT  FIXED — ROTATION.     [CHAP.  I. 

But  AC +  BC=AB.    Hence 

AC—        <"3  •  ^-Bsin  tj  „^  _        00,  .AB sin  a,  ,„, 

(Uisinai  +  ftjasinaj'  ajisinai  +  (Wasinaa 

We  thus  know  the  position  of  the  resultant  go,-  in  the  plane  of 
ooi  and  ooj. 

Magnitude  and  Direction  of  the  Resultant.* — If  we  lay  off  in 
Fig.  3,  ftJi  and  ooi,  then,  just  as  for  linear  velocities,  OC  =  oor  gives 
the  magnitude  and  direction  of  the  resultant. 

Take  rectangular  axes  OX,  OY,  Fig.  1,  in  the  plane  of  go,,  go2, 
and  let  OX  be  parallel  to  AB.  Let  go,  make  the  angle  a,  with  OX 
and  /3i  with  OY,  and  oo^  make  the  angle  a^  with  OX  and  fJ-i  with 
OY.  Denote  the  algebraic  sum  of  the  components  parallel  to  OX 
by  oox  and  parallel  to  OF  by  ooy.    Then  we  have 

cox  =  GOi  cos  a,  +  GOi  COS  aa;  | 

GOy  =  COi  COS  pi  +  CJa  COS  Pa;  J 

where  we  must  pay  regard  to  signs.  Thus  components  in  the 
direction  OX,  O  Y  are  positive,  in  the  opposite  directions  negative. 

If  the  resultant  oor  makes  the  angles  d  and  e  with  the  axes  of  X 
and  Y,  we  have 

cos  d  —  —  cos  e  =  -^ (4) 

Squaring  and  adding, 

GOr  =   ^  QOx    +  Oiiy (5) 

The  magnitude  and  direction  of  the  resultant  are  thus  deter- 
mined. 

Also  if  ^1  is  the  angle  of  go,  with  the  resultant,  and  %  the  angle  of 
GiJa  with  the  resultant,  and  &  the  angle  between  w,  and  oot ,  we  have 
directly  from  Fig.  2 

sin  01  =  —  sin  0,    sin  Oj  =  —  sin  0, (6) 

and 

OOr  =    \/cOi^  +  CJa*  ±  200\00i  COS  9, (7) 

where  the  ( + )  sign  is  used  when  Q  is  less  than  90°,  and  the  (— )  sign 
when  0  is  greater  than  90°. 

The  tangent  of  the  angle  d  which  the  resultant  makes  with  AB 
or  OX  is 

tand  =  -'' (8) 

GOx 

From  (6)  and  (7)  we  can  find  the  magnitude  and  direction  of  the 
resultant  directly  if  6i  is  known.    If  a,  and  aa  are  known,  (3)  and  (5; 
give  wr ,  and  (4)  or  (8)  the  direction. 
From  (1)  we  have  also 

GOr 

where  regard  must  be  had  for  the  signs  of  oo,p,  and  oo^p-i  in  any 
case.  From  (9)  for  any  given  point  P  for  which  p,  and  p^  are 
known,  we  can  locate  the  resultant  by  describing  a  circle  with 
centre  P  and  radius  pr  and  drawing  GOr  tangent  to  this  circle  in  the 
direction  given  by  (6). 

*  Compare  Statics. 


CHAP.   I.]    TWO    PARALLEL   AXGULAR   DISPLACEMENTS,  ETC. 


181 


The  same  formulas  hold  good  for  two  concurring  angular  ac- 
celerations. We  have  only  to  replace  go  by  a.  [The  student  will  of 
course  not  confuse  this  a,  which  stands  for  angular  acceleration, 
with  ci-i ,  cXi  in  the  formulas,  which  stand  for  angles.] 

The  same  formulas  hold  good  also  for  two  concurring  component 
angular  displacements.    We  have  only  to  replace  cj  by  d. 

When  the  Angular  Displacements,  Velocities  or  Accelerations  are 
Parallel. — In  this  case  ai  and  aa  are  equal,  S  =  0,  the  intersection  O 
is  at  an  infinite  distance,  o^r  =  <»i  +  oja ,  and  we  have  from  (,2) 


AC=~  .  AB,    BC=~.AB; 


0) 


and  hence,  multiplying  the  first  by  ooi  and  the  second  by  goq  , 

Goi  .  AC  =  &3q  .  BC,    or    —  =  -—, 

002      AC 

To  prove  this  independently,  take  C  as  the  point  of  moments 


(2) 


Then  whether  the  line  representatives  act  in  the  same  or  in  opposite 
directions,  we  have 

GO,pi  —  GOiPi  =  0,       or       GOipi  =  COiPi. 

But  from  similar  triangles 

»,      AC     ^  ooi      BC 

—  =  ^-,    hence    —  =  ^^. 
Pi      BC  coi      AC 

The  same  holds  for  angular  displacements  or  accelerations. 

We  see  from  (1)  that  the  distances  AC  and  BC  depend  only  upon 
the  magnitudes  of  goi  and  co^  and  the  distance  AB,  and  not  at  all 
upon  the  common  direction  of  ooi  and  002.  Therefore  if  ooi  and  <»j 
always  pass  through  the  points  A  and  B  no  matter  what  their 
common  direction,  the  resultant  aor  always  passes  through  C.  The 
point  C  is  then  the  point  of  application  of  the  resultant  cor  for  aU 
directions.  .  1       j- 

Hence,  the  resultant  of  tivo  parallel  component  angular  dis- 
placements,  velocities  or  accelerations  is  in  their  plane  and  equal 
in  magnitude  to  their  algebraic  sum.  It  acts  parallel  to  the  com- 
ponents in  the  direction  of  the  greater.  If  the  components  always 
pass  through  tivo  given  points  A  and  B,  the  resultant  always  passes 
through  a  point  C  no  matter  tvhat  the  common  direction.  This 
point  C  is  then  the  point  of  application  of  the  resultant.  It  is  on 
the  straight  line  AB  or  this  line  produced,  and  dimdes  it  into  seg- 
ments inversely  as  the  components.  Or  tlie  products  of  the  com- 
fonents  into  their  adjacent  segments  are  equal.  (Compare  btatics— 
'arellel  Forces.) 


182        RIGID  SYSTEM  WITH  ONE  POINT  FIXED — ROTATION.    [CHAP.  I, 

Cor.  1.  When  the  components  act  in  the  same  direction,  the 
resultant  lies  within  the  components  and  nearest  the  larger.  When 
the  components  act  in  opposite  directions,  the  resultant  lies  without 
the  components  and  on  the  side  of  the  larger. 

CoR.  2.  When  the  components  are  opposite  and  equal  in  magni- 
tude, oor  =  0.  Also  from  (1),  AC  =ao,  BC  =  ao,  or  the  resultant  is 
zero  and  acts  at  an  infinite  distance. 

That  is,  equal  and  opposite  parallel  components  cannot  have  a 
single  resultant. 

Such  a  system  is  called  a  coapl6.  (Compare  Statics— Parallel 
^Forces.) 


EXAMPLES. 

tl)  A  rigid  system  has  two  component  rotations  of  2  and  4 
radians  about  axes  inclined  60°.     Find  the  resultant  rotation. 

Ans.  Component  rotations  are  understood  to  be  simultaneous  unless  other- 
wise specified  (page  172).  Hence  magnitude  of  resultant  rotation  is  2  4/7 
radians;  axis  inclined  at  an  angle  with  the  greater  component  whose  sine  is 

4/3 
2|/7' 

(2)  A  sphere  with  one  of  its  superficial  points  fixed  has  two  com- 
ponent rotations — one  of  8  radians  about  a  tangent  line  and  one  of 
15  radians  about  a  diameter.  Find  the  axis  of  the  resultant  dis- 
placement and  the  number  of  complete  revolutions  made  about  it. 

Ans.  Inclination  of  axis  to  greater  component  at  an  angle  whose  tang  is 

—r.     Resultant  displacement  17  radians,  number  of  complete  revolutions  ;r-. 
15  27t 

(3)  A  sphere  is  rotating  uniformly  about  a  diameter  at  the  rate 
of  10  radians  per  min.  Find  (a)  the  component  angular  velocity 
about  another  diameter  inclined  30°  to  the  former,  and  (b)  the  com- 
ponent rotation  produx;ed  in  2  min.  about  a  diameter  inclined  45°  to 
the  first. 

Ans.  (a)  5  4/3  radians  per  min. ;  (6)  10  4/2  radians, 

(4)  A  pendulum  suspended  at  a  point  in  the  polar  axis  of  the 
earth  oscillates  in  a  vertical  plane.  Find  the  motion  of  this  plane 
relative  to  the  earth. 

Ans.  The  plane  of  the  pendulum  is  fixed  in  space,  and  the  motion  of  the 
earth  with  reference  to  this  plane  is  a  rotation  from  west  to  east  at  the  rate  of 
one  revolution  per  day.  The  motion  of  the  plane  relative  to  the  earth  is  then 
from  east  to  west  at  the  same  rate  of  one  revolution  per  day. 

(5)  A  pendulum  is  hung  at  a  place  of  latitude  A  and  oscillates  in 
a  vertical  plane.  Find  (a)  the  angular  velocity  of  the  plane  of  the 
pendidums  motion  relative  to  the  earth,  and  (6)  the  time  in  xvhich 
this  plane  ivill  make  one  complete  revolution  at  a  place  in  latitude 
60°  iV. 

Ans.  The  angular  velocity  of  the  earth  about  its  axis  is  27f  radians  per  day. 
The  component  of  this  in  the  direction  of  an  axis  through  the  centre  of  the 
earth  and  the  point  of  suspension  of  the  pendulum  is  2n  sin  A  radians  per  day 
from  west  to  east.  This  is  the  motion  of  the  earth  relative  to  the  plane  of  the 
pendulum.     Hence — 

(tl)  The  motion  of  the  plane  of  the  pendulum  relative  to  the  earth  is  2it  sin  \ 
-radians  per  day  from  east  to  west; 


-CHAP.  1.]  EXAMPLES.  183 

(6)  The  time  of  revolution  is  — — —  -      ^      -  _?_  d„v8 
aff  sin  A       sin  60  ~    ^/^    *^®' 

(6)  A  cube  rotates  about  a  vertically  upward  axis  through  one  of 
tts  edges.  At  a  given  instant  at  which  the  diagonal  of  the  upper 
surf  ace  passing  through  the  axis  points  north  the  cube  has  an  an- 
gular velocity  of  iO  radians  per  sec,  and  begins  to  have  a  uniform 
angular  acceleration  about  an  axis  vertically  dowmvards  through 
the  same  edge  of  6  rad.-per-sec.  per  sec.  Find  (a)  the  direction  in 
which  the  diagonal  will  point  after  20  sec. ;  (6)  the  number  of  revolu- 
tions made  by  the  cube. 

Aus.  From  the  equations  of  motion  page  73  we  have  co  =  coi  —  at,  6  —  Qi 
=  ooit  —  2^^^-  ^®  ^^v®  '"i  =  40  radians  per  sec.,  a  =  6  rad.-per-sec.  per 
sec,  t  =  20  sec,  8i  =  0. 

GO  ^0 

In  the  time  ti  =  —  =  ^  sec.  the  cube  comes  to  rest  and  has  the  angular 
displacement  9i  =  -^  radians  towards  the  east. 

o 

It  then  moves  in  the  opposite  direction  towards  the  west  durinff  the  time 
.       «n      20       40  ,       ,  ,  ^1 

t  =  aO  — Q   ~  IT  ^®^-  ^^"  undergoes  the  angular  displacement  0  =  —at^  = 

—^  radians.     Hence  (a)  the  angular  displacement  from  the  north  point  to- 

Avards  the  west  is  400  radians  or  63.661  revolutions,  or  63  revolutions  and 
26TM  W.     The  direction  of  the  diagonal  is  then  S.  57°.96  E.     (b)  The  total 

2000 
angular  displacement  is  — ^ —  radians,   hence  the  number  of  revolutions  is 

333.3 


(7)  A  sphere  is  rotating  at  a  given  instant  about  a  given  diameter 
ACB  with  an  angular  velocity  of  4  rad.  per  min.  It  has  an  angular 
acceleration  of  2  rad.-per-min.  per  min.  about  a  diameter  inclined 
30°  to  ACB.  Find  (a)  the  angular  velocity,  and  (6)  the  angular  dis- 
placement after  20  min. 

Ans.  (Page  174.)  (a)  4v  101  +  10  ^S  rad.  per  min.  inclined  to  CB  at  an 

angle  whose  tangent  is -=  ;  (b)  80  y  26  +  5  |/3   radians  inclined  to  CB 

5 

at  any  angle  whose  tangent  is -=. 

<&  -|-  o  y  o 

(8)  A  rigid  sy stein  has  one  point  fixed.  The  co-ordinates  of  this 
Xyoint  xvith  reference  to  any  point  of  the  system  taken  as  origin  are 
at  any  instant  x=  +B  ft.,  y  =  +  4  ft.,  z  =  0.  The  component  an- 
gular velocities  at  this  instant  are  sji  =  40,  odi  ■=  50,  oo^  =  60  radians 
X)er  sec,  the  line  representatives  making  the  angles  ai  =  60',  /^i  =  150% 
r.  =  90°;  a2  =  120%  ^^  =  30%  r^  =  90';  as  =  120%  ^3  =  150%  r^  =  90\ 
Find  the  resultant  angular  velocity. 

Ans.  (See  Example  (1),  page  67.)  The  component  angular  velocities  are  in 
one  plane  and 

fija;  =  —  35  radians  per  sec,     GOy=—  43.3  radians  per  sec. 

The  resultant  is  oo,-  =  55.67  radians  per  sec,  its  line  representative  or  the 
instantaneous  axis  of  rotatioii  making  with  the  horizontal  the  angle  d  —  128° 
67'  17",  and  with  the  vertical  the  angle  e  =  141°  2'  43".    If  we  look  along  this 


184  KINEMATICS  OF  A  RIGID  SYSTEM — ROTATION.        [CHAP.  I, 

line  representative  wliicli  passes  through  the  fixed  point,  towards  the  origin, 
the  rotation  will  be  seen  as  counter-clockwise. 

The  moment  of  the  resultant  angular  velocity  oor  with  reference  to  the 
point  0  gives  us  the  linear  velocity  of  rotation  at  0  about  the  instantaneous 
axis,  «j,  =  -f  10  ft.  per  sec.  in  a  direction  through  0  perpendicular  to  the  plane 
XT,  or  along  Z,  from  0  towards  Z. 

The  distance  of  0  from  the  axi^  is  p  =  about  0.18  ft. 

The  equation  of  the  axis  is  y  =  1.237a!  -|-  0.286.  Its  intercepts  on  the  axis 
are  /  =  +  0.286  ft.,  x  -  -  0.232  ft. 

(9)  Express  and  solve  the  same  example  for  component  angular 
accelerations  and  displacements. 

(10)  A  rigid  system  has  one  point  fixed.  The  co-ordinates  of  this 
point  with  reference  to  any  point  of  the  system  taken  as  an  origin 
are  at  any  instant  x  =  +  3  ft.,  y=+  ^ft.,  z=  +5ft.  The  compo- 
nent angular  velocities  at  this  instant  are  goi  —  40,  cot  =  50,  oo-i  —  60 
radians  per  sec.,  the  line  representatives  making  the  angles  with  the 
axe4  o-j  =  60°,  (ii  =  100°,  yi  obtuse ;  a^  =  100°,  /Sj  =  60°,  y^  acute;  as 
=  120°,  fJa  =  100%  ya  acute.    Find  the  resultant  angular  velocity. 

Ans.  (See  Example  (2),  page  67.)    We  have 

oox=-  18.6824,     GOy=-{-  7.635,     ooz  = -{-  59.391  radians  per  sec. 

The  resultant  angular  velocity  is  cor  =  62.73  radians  per  sec,  its  line  rep- 
resentative making  with  the  axes  the  angles 

d  =  118°  17'  33",    e  =  85°  6'  12",    /  =  13°  30'  24". 

This  line  representative  passes  through  the  fixed  point  and  gives  the  in- 
stantaneous axis  of  rotation.  If  we  look  along  this  line  towards  the  origin,  the 
rotation  will  be  seen  as  counter-clockwise. 

The  velocities  of  rotation  at  0  along  the  axes  are 

rz  =  +  97.6346,    vx  = -^  199.389,    %  =  -  371.585  ft.  per  sec. 

The  resultant  velocity  of  rotation  at  0  is  «r  =  407.6  ft.  per  sec,  making 
•with  the  axes  angles 

a  =  60°  42'  57",    b  =  131°  46'  34",    c  =  76°  8'  31". 

The  equations  of  the  projections  of  the  axis  upon  the  co-ordinate  planes  are: 

on  plane  XT,    y  =  -  0.408a!  -f  5.226 ; 
"      '•      TZ,    s  =+7.778^^-26.115; 
"      "     ZX,    x=  -  0.314s  -\-  4.572. 

The  axis  pierces  the  plane  XT    at  «'  =  +   4.572  ft.,    /  =  +   3.357  ft.; 
"      "        '=        "       "      TZ     "  y' = -f    5.226  ft.,     2'  =  - 14.56  ft.; 
«•       ••        "        "       "      ZX      "  s'  =- 26.115  ft.,     «' =  +  12.809  ft. 

(11)  Express  and  solve  the  same  example  for  component  angular 
accelerations  and  displacements. 

(12)  Let  the  axes  of  tivo  concurring  angular  velocities  of  a  mgid 
sy  stern,  gdi  =  20,  ooi  =  30  radians  per  sec,  pass  through  the  points  A, 
B  of  the  system,  the  distance  AB  =  2  ft.,  and  the  angles  a^  =  60°,  a^ 
=  30°.  Find  the  point  C  on  the  line  AB  through  which  the  resultant 
axis  passes,  and  the  magnitude  and  direction  of  the  resultant  angu- 
lar velocity. 


CHAP.  I.]  EXAMPLES.  185 

Ans.  AC  =  0.928  ft.,  ajx  =  +  15.98  radians  per  sec,  GOy=  —  32.33  radians 
per  sec.     The  angle  of  the  resultant  with  AB  is  given  by 

32  32 
tan  d  =  -  j^  =  -  2.022,     or    d  =  63°  41'  =  BCcor. 

The  resultant  is 

GOr  =  36.05  radians  per  sec. 

We  have  also  for  the  angle  of  the  resultant  with  t»i ,  since  6  =  90", 
on 

8i»  6,  =  sl^  =  0.832,    or    6,  =  56°  19'. 
00.05 

(13)  Express  and  solve  the  same  example  for  component  angular 
accelerations  and  displacements. 


CHAPTEE  IL 
MOMENT  OF  A  COUPLE. 


DISPLACEMENT  OP  A  RIGID  SYSTEM.  RIGID  PLANE  SYSTEM.  COMPOSITION 
AND  RESOLUTION  OF  TRANSLATION  AND  ANGULAR  DISPLACEMENT.  COM- 
POSITION AND  RESOLUTION  OP  TRANSLATION  AND  ANGULAR  VELOCITY. 
CENTRAL  AXIS.  SCREW  MOTION.  ROTATION  AND  RECTILINEAR  TRANS- 
LATION. COMBINED  PARALLEL  ROTATIONS,  ONE  AXIS  FIXED.  INTER- 
SECTING AXES,  ONE  AXIS  FIXED.  ANALYTIC  DETERMINATION  OF  RE- 
SULTANT ANGULAR  VELOCITY  AND  VELOCITY  OF  TRANSLATION  FOR  NON- 
CONCURRING  ANGULAR  VELOCITIES. 

Moment  of  a  Couple.* — We  have  just  seen  in  the  preceding 
Chapter,  page  181,  that  two  parallel  equal  and  opposite  components 
acting  at  different  points  of  a  rigid  system  constitute  a  couple.  We 

may  have  then  an  angular-displace- 
ment couple  or  angular- velocity  cou- 
ple or  angular-acceleration  couple. 

Let  +  00,  —  CO,  acting  at  the  points 
A,  -B  of  a  rigid  system,  constitute  an 
~Q,      angular-velocity  couple. 

If  we  take  any  point  C  between 
the  components,  or  any  point  d ,  d 
on  either  side,  in  the  plane  of  the 
components,   we    have    in  the   first 


Ci 


case,  denoting  the  distance  AB  by  p,  for  the  moment  about  C,  just 
as  for  linear  velocities  (page  60), 

-a>.AC-oo.BC=-  oo{AC  +  BC)  =  -poo. 

In  the  second  case,  for  the  moment  about  Ci  we  have 

00  .  CiA  -  00.  aB  =  -  co{aB  -  C^A)  =  -poo. 

In  the  third  case,  for  the  moment  about  d  we  have 

—  CO .  CiA  +  00 .  CiB  =  —  Go{CiA  —  CiB)  =  —  pan. 

Hence  the  moment  about  every  point  in  the  plane  of  the  couple  is 
constant  and  equal  to  ±  poo,  the  (-1-)  or  (— )  sign  denoting  direction 
just  as  for  moment  of  linear  velocity  (page  60). 

For  an  angular-acceleration  couple  we  have  in  the  same  way 
±  pcx,  for  an  angular-displacement  couple  ±jp6. 

We  see  then  that  the  moment  of  a  couple  is  the  same  for  every 
point  in  its  plane  and  equal  to  the  product  of  either  of  the  compo- 
nents by  the  distance  between  them. 

*  Compare  Statics— Parallel  Forces. 

186 


CHAP,  n.]  DISPLACEMENT   OF   A   RIGID    SYSTEM. 


187 


a.-*^' 


Composition  and  Resolution  of  Translation  and  Angular  Displace- 
ment.—Let  a  rigid  system  have  a  rotation  of  OiOO^  =  Q  radians 
about  an  axis  AOB  through  the 
point  O,  and  AB  =  9  be  the  line 
representative. 

If  we  take  any  other  point  of 
lae  system,  as  Oi ,  and  at  this  point 
apply  the  two  equal  and  opposite 
angular  displacements  Oia  —  —B 
and  Oih  =  +  6,  both  parallel  to 
AB,  it  is  evident  that  the  motion 
of  the  system  is  not  affected.  We 
have  then  the  angular  displace- 
ment about  the  axis  AOB  reduced  to  an  equal  angular  displace- 
ment 0\b  about  a  parallel  axis  through  Oi  and  a  couple  AB  and 
Om. 

The  moment  of  this  couple  is  the  same  for  every  point  in  its 
plane  and  equal  to  pO,  where  p  is  the  perpendicular  distance  be- 
tween the  components  AB  and  Om  of  the  couple. 

But  we  have  seen  (page  177)  that  the  moment  pO  corresponds  to 
a  linear  displacement  in  a  plane  perpendicular  to  the  plane  of  the 
couple  of 


T  =  2p  sin  -, 
making  an  angle  00x0%  withjp  given  by 

Tt—B 


00.0%  = 


(1) 


(2) 


Hence,  an  angular  displacement  Q  about  any  given  axis  can  be 
resolved  into  an  equal  angular  displacement  aboid  a  parallel  axis 
through  any  point  of  the  system  and  a  linear  translation  in  the 
plane  of  rotation  of  the  system  whose  magnitude  and  direction  are 
given  by  (1)  and  (2). 

Conversely,  the  resultant  of  the  rotation  of  a  rigid  system  about 
a  given  axis  and  a  translation  in  any  given  direction,  is  an  equal 
rotation  about  a  parallel  axis,  whose  position  with  reference  to  the 
first  can  he  determined  by  (1)  and  (2). 

Cor.  1.  Two  non-concurring  angular  displacements  can  be  re- 
duced to  a  resultant  angular  displacement  about  a  resultant  axis 
at  any  point  and  a  couple  which  causes  translation. 

Cor.  2.  Hence  if  we  have  any  number  of  component  angular 
rotations  about  any  axes,  whether  these  axes  intersect  or  not,  we 
can  reduce  each  to  an  equal  rotation  about  a  parallel  axis  through 
some  one  point  of  the  system  and  a  translation  of  the  system. 

The  resultant  translation  can  then  be  found  as  on  page  35,  and 
I  he  resultant  rotation  as  on  page  173,  for  simultaneous  angular  dis- 
placements. 

Cor.  3.  Therefore  any  number  of  component  translations  and 
rotations  can  all  be  reduced  to  a  single  translation  and  a  single 
rotation  about  any  given  point.  It  is  evident  that  this  single  rota- 
tion is  not  affected  by  the  position  of  the  point,  which  affects  the 
translation  only. 

Displacement  of  a  Rigid  System. — Any  displacement  of  a  rigid 
£ystem  may  be  produced  by  a  translation  and  an  angular  displace- 
ment. 


188 


RIGID  SYSTEM— TRANSLATION  AND  ROTATION.     [CHAP.  II» 


•C. 


Let  Ai,Bi,  Ci  be  the  positions  of  any  three  points  which  determine 

the  initial  position  of  the  rigid  system. 

Let  A-2 ,  Bi,  Ci  be  the  final  position  of 

these  jjoints  after  any  displacement. 

First  let  the  system  be  translated,  sa 

that  Ai  comes  to  its  final  position  Aj. 

Then  Bi  and  Ci  will  take  the  positions 

b  and  c,  the  lines  Bib  and  CiC  being 

equal  and  parallel  to  AiAi.     We  see 

then  that    Ai  is  a  fixed  point  in  the 

system  so  far   as   the   two    positions 

AiBiCi ,  AiBiCi  are  concerned. 

But  we  have  seen  (page  173)  that  in  every  possible  displacement 

of  a  rigid  system  with  one  point  fixed  there  is  an  axis  of  rotation 

fixed  in  the  system  which  remains  unchanged.     Hence  A^cb  can  be 

brought  to  the  position  A^BiCi  by  rotation  about  that  axis. 

OoK.  i.  It  follows  that  the  displacement  of  a  rigid  system  is 
known  if  the  magnitude  and  direction  of  the  linear  displacement  of 
any  point  is  known,  and  also  the  magnitude  and  direction  of  the 
angular  displacement  of  the  system  about  that  point. 

Cor.  2.  Also,  the  displacement  of  a  rigid  system  is  known  if  the 
magnitude  and  direction  of  the  component  linear  displacements  of 
any  point  parallel  to  three  rectangular  axes  and  of  the  component 
angular  displacements  of  the  system  about  axes  parallel  to  the  first 
through  the  point  are  known. 

Rigid  Plane  System. — Any  displacement  of  a  rigid  plane  system  in 

its  own  plane  may  be  produced  by  rotation  about  some  point  in  the  plane. 

Let  AiBi  and  A-^B^  be  the  initial  and  final  positions  in  the  plane, 

of  the  same  line  of  the  system,    o  ^ 

so  that  AiBi  and  A^Bi  are  of         \\ns.   -^-  -,r 

equal  length.    Join  AA-i  and         w  N^^  ^---  ^^-  a 

BiBi  by  lines  and  bisect  these 
lines  at  C  and  D.     Erect  per- 

Eendiculars  at    the    points  of  \    \      \  ^"-x^  7d 

isection  C  and  D  and  produce 
them  to  intersection  at  O. 
Then  by  construction  OAi  — 
OAi,  and  0J5i  =  OB^,  and  A,B, 
=  AiBi.  Hence  the  two  triangles  OAiBi  and  OA^B^  are  in  all 
respects  equal  and  the  line  AiBi  may  be  brought  to  coincide  with 
AiBi  by  rotation  about  the  point  O. 


If  AiBi  is  parallel  to  AiBi ,  we  have  translation  only  and  the 
point  O  is  at  an  infinite  distance. 


CHAP.  II.]  DISPLACEMENT   OF   A    RIGID   SYSTEM.  189 

Since  the  angle  A.OBi  —  the  angle  A-^OB^ ,  if  we  take  the  angle 
A^.OBi  from  both  we  have  AiOA^  =  B1OB2.  If  then  the  displace- 
ment is  such  that  Ai  falls  on  OBi  or  on  OBi  produced,  Ai  must  be 
on  OB 2  or  OB 2  produced. 

In  both  cases  OC  and  OD  coincide  and  do  not  intersect,  but  it  is 
evident  that  in  such  case  the  point  O  in  which  AiB,  and  A-.B^.  inter- 
^sect  is  the  point  about  which  rotation  would  produce  the  given  dis- 
placement. 

If  in  any  case  A^A^  and  B1B2  are  indefinitely  small,  the  point  O 
is  called  the  instantaneous  centre  of  motion. 

Any  Displacement  of  a  Rigid  System. — Any  displacement  of  a 
rigid  system  may  be  produced  by  rotation  about  an  axis  and  a  trans- 
lation in  the  direction  of  that  axis. 

Let  AB  and  BC  represent  the  resultant 
translation  and  rotation  to  which  the  compo- 
nent translations  and  rotations  of  the  system 
can  be  reduced  (page  187,  Cor.  3). 

Draw  AD  and  DB  parallel  and  perpendic- 
ular to  BC.  Then  the  translation  AB  is  re- 
solved into  the  two  components  AD  and  DB. 

But  the  resultant  of  DB  and  BC  (page  187)  is  an  equal  rotation 
about  an  axis  parallel  to  BC.  Hence  the  ti'anslation  AD  and  the 
rotation  BC  are  reducible  to  the  translation  AD  and  a  rotation 
about  an  axis  parallel  to  AD. 

Composition  and  Resolution  of  Translation  and  Angular  Ve- 
locity or  Angular  Acceleration.  —  Let  a  rigid  system  have  an 
\     angular  velocity      (      ^    \         ^radians  per  sec.  ( about  an 

I  angular  acceleration  )  (a  radians-per-sec.  per  sec.  ) 

axis  AOB  through  the  point  O,  and 

let  AB  —  \'\''^ihe  the  linear  rep- 

resentative. 

If  we  take  any  other  point  of  the 
system,  as  O. ,  and  at  this  point 
apply  two  equal  and  opposite 
3  angular  velocities  )  0  a  = 
\    angular  accelerations  ( '       ' 

^  |i:[andO,6=j+:f,itisevi- 

*  *"'•'  dent  that  the  motion  of  the  system 

is  not  affected.    We  have  then  the  ]  angular^accelemtion  [  ^^"* 

the  axis  AOB  reduced  to  an  equal  j  angular^  acceleratfon  [  ^'^ 
about  a  parallel  axis  through  Oi  and  a  couple  represented  by  AB 
and  da. 

The  moment  of  this  couple  is  the  same  for  every  point  m  its 

plane  (page  186)  and  equal  to  ]  p^  [  »  where  p  is  the  perpendicular 
distance  between  the  two  axes. 

But  we  have  seen  (page  177)  that  the  moment  j  ^^  [  gives  the 

linear  j  acceleration  f\^^^  direction  perpendicular  to  the  plane  of 
the  couple.  Since  the  moment  of  the  couple  is  the  same  for  every 
point  in  its  plane,  we  have  then  translation  of  the  entire  system  m 
a  direction  perpendicular  to  the  plane  of  the  couple,  as  well  as 
simultaneous  rotation  about  the  axis  through  Oi.    The  direction 


a-<: 


190  RIGID  SYSTEM — TRANSLATION  AND  ROTATION.    [CHAP.  11, 

and  magnitude  of  this  translation  will  depend  upon  the  point  Oi , 
but  the  rotation  will  be  the  same  wherever  the  point  Oi  may  be 
taken. 

Hence,*  an  \  ang^laraccSicm  \  «/  «  ^^^^  'y'*^^  «^«^^  «^^ 
aads  can  he  resolved  into  an  equal  \  angZalZcTeferaficm  \  «^«^^*  « 
parallel  axis  at  any  distance p,  and  a  ]  fjp^pj^fJ^fjfjf.  [  of  translation 
)  /  ^  '^  X  *w  a  direction  at  right  angles  to  the  plane  of  the  axes. 

c'onvLely,  m  re^yltant  of  an  j  ^^"^^^Jf^T^i'Ln  .  i  »/ " 
rigid  system  about  a  given  axis  and  a  \  rif^ti^?,^fi(y~,  [  Of  translation 

\j\in  any  direction  is  an  equal  \  ^^^-^i'Lc^^^^^^     \  «^«-*  « 

f  V  1 

p  =  - 

00 

parallel  axis  distant  \  .  >  ina  direction  perpendicular  to  the 

[  "  J 

plane  of  \  ^i  and  the  given  axis. 

This  parallel  axis  is  the  instantaneous  axis. 

CoR.  1.  Hence  if  we  have  any  number  of  component  ]  anfular 
accelerations  [  ^^*^^*  ^^y  axis,  each  can  be  reduced  to  an  equal 

point  of  the  system,  and  a  \  acceleratTon  [  °^  translation  of  the 
system.    We  can  then  find  the  resultant  \  acceleration  C  ^^  trans- 

lation  as  on  page  {  g  [  and  the  resultant  |  a„*"uCLoetrlS,„  (  "« 
on  page  176. 

Cor  2.  Therefore  any  number  of  component  ]  o!!f!!lar  ac- 
celSfons  ( ^^d  ]  acletriifons  \  «*  translation,  can  all  be  reduced 
to  a  single  resultant  j  onceleratwn  1  °^  translation  and  a  single  re- 
«^t^^*  1  aS^i^trrnSfon  h^«^t  ^^  ^^^  ^^^^^^^h  any  one  point 
of  the  system.  The  j  acceleration  [  ^^  translation  will  vary  in  di- 
rection and  magnitude  with    the    point    chosen.     The  ]  anfuKr 

aopplpration  [  ^^^^  ^®  ^he  same  no  matter  what  point  is  chosen. 

Central  Axis. — Any  number  of  component  angular  velocities  of 
a  rigid  system  can  be  reduced  to  a  single  angular  velocity  about  a 
determinate  axis  and  a  simultaneous  velocity  of  translation  of  the 
system  along  that  axis.  Such  an  axis  is  called  the  central  axis,  and 
such  motion  is  called  screw  motion. 

*  Compare  Statics — Non-concurring  Forces. 


CHAP.  II.] 


CENTRAL   AXIS. 


191 


Thus  let  OA  and  OC  represent  the  resultant  linear  velocity  of 
translation  Vr  and  the  resultant  angu-  r^ 

lar  velocity  cjr,  to  which,  as  we  have 
just  seen,  all  the  angular  velocities  can 
be  reduced. 

Draw  AD  and  OD  parallel  and  per- 
pendicular to  OC.  Then  the  velocity  of 
translation  OA  =  Vr  is  resolved  into  the  two  components  AD  and 
OD.  But  the  resultant  of  OC  and  OD  is  an  equal  angular  velocity 
about  an  axis  parallel  to  OC  (page  190). 

Hence  the  velocity  of  translation  OA  =  Vr  and  the  angular  ve- 
locity OC  =  oor  are  reducible  to  an  equal  angular  velocity  about  an 
axis  parallel  to  OC  and  a  linear  velocity  of  translation  AD  along 
that  axis 

This  axis  is  called  the  central  axis,  and  may  be  located  by  the 
following  geometric  construction. 

At  any  point  O  of  the  system  taken  arbitrarily  let  the  velocity 

of  translation  be  V,-  and  the  ro- 
tation axis  through  O  be  oor, 
making  the  angle  (p.  Through 
O  draw  a  line  OD  =p  perpen- 
dicular to  the  plane  of  Vr  and 
GOr,  SO  that  poor  =  Vr  siu  0,  or 
Vr  sin  <p 

p  = 

OJr 

Then  a  line  through  D  parallel 
to  the  rotation  axis  at  O  will  be  the 
central  axis.     (Compare  Statics— Non-concurring  Forces. 

Screw  Motion. — Let  u,-  denote  the  resultant  velocity  of  trans- 
lation along  the  central  axis.  This  is  called  the  velocity  of  advance. 
The  distance  d  advanced  during  one  complete  rotation  of  the  sys- 
tem is  called  the  pitch  of  the  screw,  and  the  distance  advanced  dur- 
ing a  rotation  of  one  radian,  or  —,  we  call  the  unit  pitch  of  the 

screw.  •       .,      -•          £ 

If  GOr  is  the  resultant  angular  velocity  of  rotation,  the  tmie  of  a 

2^ 


complete  rotation  is  ^  =  — 

00  r 

We  have  then  for  the  value  of  the  pitch 


d  =  Vrrt  = 


and  for  the  unit  pitch 


2itUr 


hence     w  = 


Ord 

"2^' 


(1) 


(2) 


A— — 

2it        oor 

If  r  is  the  radius  vector  of  any  point  of  the  system,  then  the 
linear  velocity  of  that  point  due  to  rotation  about  the  axis  is 

v  =  roor (^) 

in  a  direction  perpendicular  to  the  plane  of  the  central  axis  and  the 
radius  vector. 

The  resultant  velocity  at  that  point  is  then 


Or  =   {/u,^  +  ^''  =  "^'•y    ^  +  ^ 


(4j. 


192 


EIGID  SYSTEM — TRANSLATION"  AKD  EOTATION.    [CHAP.  II. 


The  inclination  of  the  path  at  that  point  to  the  plane  of  rotation 
is  given  by 

d 
2nr' 


tan  1  =  — 


(5) 


or  the  tangent  of  the  angle  of  inclination  at  any  point  is  equal  to 
the  ratio  of  the  pitch  to  the  circumference  of  the  circle  described 
by  the  point  relatively  to  the  axis  ;  or  it  is  equal  to  the  ratio  of  the 
unit  pitch  to  the  radius  vector  of  the  point. 

Centre  of  Parallel  Angular  Velocities.* — Let  &n,  0^2,0,73,  etc.,  be 
any  number  of  parallel  angular  velocities  passing  through  the  points 
^1 ,  A^^Aa,  etc. ,  of  a  rigid  system. 


7"* X- 


Then  the  resultant  ^r  must  be  parallel  to  the  components  and 
equal  in  magnitude  to  their  algebraic  sum,  or 

OOr  =  Gl>i  +  <i52  +  003  +  .  .  .  =  'Sqo. 

Take  any  two  components  rai  and  (»2,  and  produce  the  line  A\ , 
Ai  to  intersection  K  with  the  plane  ZX.  Drop  perpendiculars  AiBi, 
AiBi  to  this  plane  and  draw  the  line  KB1B2  in  this  plane. 

Now,  from  page  181,  the  resultant  of  goi  and  G02  is  g,7i  =  gji  +  (a2 
and  its  point  of  application  is  at  A  on  the  line  AiAi ,  so  that 

ooi  _  A-iA 

GOi      AiA' 

Drop  the  perpendicular  AB  to  the  plane  ZX.  Then  we  have  by 
similar  triangles 

A,A  ^  B2B 
A,A~B,B' 

Denote  the  distance  AiBi ,  A2B2  by  yi  ,  y-i,  respectively,  and  the 
distance  AB,  or  the  ordinate  of  the  point  of  application  of  the  re- 
sultant ^1  of  coi  and  co-x,  by  yi.    Then  we  have  by  similar  triangles 


Hence 


In  the  same  way  for  three  angular  velocities,  mi  ,  &32 ,  cja ,  we  can 
■combine  the  resultant  oi  of  ooi  and  002  passing  through  A,  with  003. 

*  Compare  Statics — Parallel  Forces. 


B2B  _  2/2-2/1 

BrB       yy-yr' 

CSJl  _ 

Vl  —  ?/l'    "^    "'        <"'  +  ^-'^ 

•CHAP.  II.]  ROTATION  AND  KECTILINEAE  TRANSLATION  COMBINED.  193 

"We  thus  obtain  for  the  ordinate  of  the  point  of  application  of  the 
resultant  of  three  forces 

In  general,  then,  for  any  number  of  parallel  angular  velocities 
we  have  for  the  ordinate  y  ot  the  point  of  apphcation  of  the  result- 
ant 

—      2ayu 

»=^ « 

In  precisely  similar  manner,  if  we  denote  the  distances  AC  and 
AD  of  the  point  of^application  of  the  resultant  from  the  planes  YZ 
and  XY  by  x  and  y,  we  have 

-=^-- <^> 

-S (3) 

Equations  (1),  (2)  and  (3)  give  the  co-ordinates  of  the  point  of 
application  of  the  resultant  for  any  number  of  parallel  angular 
velocities.  This  point  is  called  the  centre  of  parallel  angular  veloc- 
ities. The  same  equations  evidently  hold  for  parallel  angular  ac- 
celerations, by  replacing  oj  by  a.  We  have  then  the  centre  of 
parallel  angular  accelerations.  > 

The  position  of  this  centre  depends  only  upon  the  magnitude  and 
position  of  the  line  representatives  and  is  independent  of  their  com- 
raon  direction. 

If  z  is  zero,  Zi,  Zt,  etc.,  are  zero,  and  the  line  representatives  all 
lie  in  the  plane  XY.  The  centre  is  then  given  by  (1)  and  (2).  If 
2  and  y  are  zero,  the  centre  is  in  the  axis  of  Xand  is  given  by  (2). 

Rotation  and  Rectilinear  Translation  Combined.  —  Let  a  rigid 
system  have  an  angular  velocity  Ob  =  go  about  an  axis  through  O, 
perpendicular  to  the  plane  of  the  paper, 
and  at  the  same  time  a  velocity  of 
translation  u  in  a  straight  line.  Then, 
as  we  have  seen,  page  177,  v  can  be  re- 
placed by  the  couple  Oa  and  IB,  and  we 
have  at  any  instant  a  resultant  rotation 
IB  —  CO  about  a  parallel  axis  though  I  at 

a  distance  OI  =  p  =  -  in  a  direction 

GO 

perpendicular  to  that  of  v.  This  axis  is  the  instantaneous  axis; 
that  is,  the  point  /at  any  distant  has  the  velocity  v  in  one  direction 
d\ie  to  translation,  and  the  velocity  v  =  poa  m  the  other  direction 
due  to  the  couple,  and  is  therefore  at  rest. 

It  is  evident  that  every  straight  line  in  the  system  parallel  to 
the  moving  axis   at  O  and    at  a  constant   distance  from  O  of 

p  =  -  becomes  in  turn  the  instantaneous  axis  when  it  arrives  at 

GO 

the  position  J  with  reference  to  O. 

Hence  when  a  rigid  system  has  a  velocity  of  translation  m  a 
straight  Hue  and  at  the  same  time  an  angular  velocity  go  about  a 
given  axis  Ob,  the  resultant  motion  of  the  system  ts  the  same  as  tf 


194 


EIGID  SYSTEM — TRANSLATION  AND  EOTATION.    [CHAP.  lU 


a  cylindrical  surface  fixed  in  the  system,  of  radius  p,  =  — ,  rolled  on^ 

GO 

a  plane  HIH  parallel  to  the  plane  of  Oh  and  v. 

The  path  described  by  any  point  in  the  axis  Ob  is  a  straight 
line.  The  path  described  by  any  point  not  in  this  axis  is  called  a 
trochoid.    The  special  form  of  trochoid  described  by  any  point  in 

the  cylindrical  surface  is  called  a  cycloid. 

Any  internal   point  describes  a  prolate 

cycloid  ;    any  external    point,  a  curtate 

cycloid. 


Curtate  Cycloid 


The  general  form  of  these  curves  is 
shown  in  the  accompanying  figures. 
A  common  illustration  of  such  motion  is  a  wheel  rolling  in  a  straight  line 


on  a  plane. 


If  the  radius  is  p,  we  have  poa—v  and  hence  —  =  p. 

00 


The  in- 


Cyclold 


stantaneous  axis  is  at  right  angles  to  the  plane 
of  the  wheel  and  passes  through  the  point  of 
contact  with  the  plane.  The  velocity  at  the 
centre  is  t,  and  at  the  opposite  extremity  of  the 
diameter  through  the  point  of  contact  2»  in  the 
direction  of  translation.  The  velocity  of  any 
point  at  a  distance  d  from  the  instantaneous  axis  is  dco  in  a  direction  per- 

pendicular  to  the  plane  of  the  instantaneous 
Prolate  Cycloid  ^^jg  ^mj  i\^q  instantaneous  radius  vector  d. 

Combined  Parallel  Rotations — One 
Axis  Fixed. — Let  a  rigid  system  rotate 
with  the  angular  velocity  ooi  about  a 
moviiig  axis  at  Oi ,  and  at  the  same 

time  let  this  axis  revolve  with  the  angular  velocity  oo^  about  a 

parallel  fixed  axis  at  Oa. 

Then,  as  we  have  seen,  page  181,  the  resultant  f.7r  is  in  the 

plane  of  the  components  ooi  and  ooi,  is  equal    in  magnitude  to 

their  algebraic  sum  and  divides  the  straight  line  joining  Oi  and 

Oi  into  segments  inversely  as  the  components.      Also  when  the 

components  act  in  the  same  direction  the  resultant  lies  within 

the    components,  and  when  in 

opposite  directions  without  the 

components  and  on  the  side  of 

the  larger. 

Fig.  1,  then,  represents  the 

case  in  which  ooi  and  oai  are  in 

the  same  direction ;   Fig.  2,  that 

in  which  £»i  and  ao-i  are   in  op- 
posite directions  and  coi  is  the 

greater ;  Fig.  3,  that  in  which  ooi  and  ooa  are  in  opposite  directions 

and  cit  is  the  greater. 

The  resultant  angular  velocity  is  in  all  cases  then  given  by 


Fig.  1. 


GOr  =  GJi  +  ftJa, 


(1) 


where  we  take  kji  and  ftjs  with  their  proper  signs. 

This  resultant  rotation  oor  takes  place  about  the  instantaneous 
axis  through  J,  so  situated  that  (page  181) 


GOi 
OOi 


10, 
10.' 


(2> 


so  that  at  any  instant  1  has  two  opposite  and  equal  linear  velocities 
and  is  therefore  at  that  instant  at  rest.  Since  then  ooi .  10 1  =  go,  .  lOi, 


CHAP.  II.]  ROTATION   ABOUT   INTERSECTING   AXES.  195 

we  see  at  once  that  in  Fig.  2,  oa,  is  greater  than  c»2,  and  in  Fig.  3,  wi 
is  less  than  Wi. 

We  have  also  as  on  page  181,  taking  moments  about  Oa  and  Oi , 


fiji .  OiOi  =  oor .  lOi,    or    lO-i  =  — .  OiOa ; 

QOr 
ftJa  .  OiOa  =  COr.  lOi,      OT      lOi  =  -—  .  OiOa. 


(3) 


All  the  lines  in  the  system  which  successively  occupy  the  position 
of  the  instantaneous  axis  are  then  situated  in  a  cylindrical  surface 

described  about  Oi  with  the  radius  lOi  =  — ^  .  OiOa;   and  all  the 

COr 

positions  of  the  instantaneous  axis  are  contained  in  a  cylindrical 

surface  described  about  Oa  with  the  radius  JOa  =  — ^  .  O1O2 

Hence  the  resultant  motion  of  a  rigid  system  which  rotates 
about  an  axis  O:  while  at  the  same  time  this  axis  revolves  about  a 
fixed  axis  Oa  is  the  same  as  if  a  cylindrical  surface  of  radius  10 1  = 

— ^  .  OiOa ,  fixed  in  the  system,  rolls  upon  a  fixed  cylindrical  surface 

oar 

of  radius  lOi  =  -^ .  OiOa. 

CO,- 

In  Fig.  1,  a  convex  cylinder  rolls  on  a  convex  cylinder;  in  Fig. 
2,  a  smaller  convex  cylinder  rolls  within  a  larger  concave  cylinder; 
in  Fig.  3,  a  larger  concave  cylinder  rolls  upon  a  smaller  convex 
cylinder. 

The  path  described  by  any  point  in  the  moving  axis  through  Oi 
relatively  to  the  fixed  axis  at  Ci  is  a  circle. 

The  path  described  by  any  point  relatively  to  the  fixed  axis  is 
called  an  epitrochoid  when  the  rolling  cylinder  is  outside  of  the 
fixed  cylinder  and  an  hypotrochoid  when  it  is  inside.  The  special 
form  of  epitrochoid  or  hypotrochoid  described  by  a  point  in  the 
surface  of  the  rolling  cylinder  is  called  an  epicycloid  when  the  roll- 
ing cylinder  is  outside  and  an  hypocycloid  when  it  is  inside  the  fixed 
cylinder.  ,  ,  .  ^, 

When  the  distance  OiOa  is  infinite  we  have  the  case  of  the  pre- 
ceding Article,  of  a  cylinder  rolling  on  a  straight  line.   In  this  case 

When  a  cylinder  rolls  externally  upon  another  of  equal  size,  the 
special  form  of  epicycloid  described  by  a  point  in  its  surface  is 
called  the  cardioid.  In  this  case  coi  and  co  are  equal  and  in  the  same 
direction. 

When  a  cylinder,  as  in  Fig.  2,  rolls  withm  a 
concave  cylinder  of  double  its  radius,  we  have 
GO,  —  2oo2.  In  this  case  each  point  in  the  surface 
of  the  rolling  cylinder  moves  to  and  fro  in  a 
straight  line,  being  a  diameter  of  the  fixed  cylin- 
der; each  point  in  the  axis  of  the  rolling  cylinder 
describes  a  circle  of  the  same  radius  as  that  cylin- 
der and  any  other  point  in  or  without  the  rolling 
cvlinder  describes  an  ellipse  of  greater  or  less  eccentricity,  having 
its  centre  in  the  fixed  axis  at  C.  This  principle  has  been  made 
available  in  instruments  for  drawing  and  turning  ellipses. 

Rotation  about  Intersecting:  Axes— One  Axis  Fixed.  —  Let  Ca 
be  a  fixed  axis  and  about  it  let  the  plane  O.CO,  rotate  with  the 


196 


RIGID  SYSTEM— TKANSLATION  AifD  ROTATION.    [CHAP.  II. 


angular  velocity  oja.  Let  COi  be  an  axis  in  the  rotating  plane,  and 
about  that  axis  let  a  rigid  system  rotate  with  the  angular  velocity 
tt»i  relatively  to  the  rotating  plane. 


If  we  lay  off  from  C  the  line  representatives  Ca  =  ooi  and  CO  = 
cBi  along  the  axes,  the  diagonal  Ci  of  the  parallelogram  gives  the 
magnitude  and  direction  of  the  resultant  angular  velocity  oor .  The 
instantaneous  axis  then  occupies  the  position  CI.  If  we  denote  the 
angles  ICd  and  ICOi  by  aa  and  a, ,  we  have 


tan  aa  = 


tan  ai  = 


GOi  sin  (oTi  +  ai) 
Oh  +  ooi  cos  («!  +  ai) 

sin  (ai  +  at) 


—  am  (ai  +  ai) 

OOi 


1  +  —  cos  (ai  +  ai) 


Oil 

1-  cos  (ai  +  ai) 

OOi 


(1) 


(2) 


From  (1)  and  (2)  we  can  find  ai  and  a» ,  when  the  angle  between 
the  axes  (ai  +a »)  and  the  angular  velocity  ratio  —  are  given.    We 

GOi 

have  also 


and 


GOr 

OOi 


GOr'  =  tai"  +  coi^  ±  2ooicOi  COS  (ofi  +  eta),      .      . 
sin  (ai  +  ai)  oor       sin  (ori  +  aa)  coi      sin  aa 


sm  «i 


Sm  rra 


sm  a  I 


(3) 


(4) 


All  lines  which  come  successively  into  the  position  of  the  in- 
stantaneous axis  are  in  the  surface  of  a  cone  described  by  the  revo- 
lution of  CI  about  COi ;  and  all  the  positions  of  the  instantaneous 
axis  lie  in  the  surface  of  a  cone  described  by  the  revolution  of  CI 
about  COi. 

Hence  the  motion  of  the  rigid  system  is  such  as  would  he  pro- 
duced by  the  rolling  of  the  cone  CIOi ,  fixed  in  the  system,  about  the 
fixed  cone  Cld. 

If  ra  is  the  radius  dl  of  the  fixed  cone,  and  ri  the  radius  Od  of 
the  rolling  cone,  we  have 


rict>i  =  raoaa ,   or   — 


Vi 

ri 


For  the  height  Cd  =  hot  the  fixed  cone  we  have 

ri  +  ra  cos  (ai  +  as) 


hi  —  Vi  cotang  era  = 


sin  (ai  +  as) 


(5) 


(6) 


CHAP.II.]  ANALYTICAL  DETERMINATION  OF  RESULTANT  VELOCITY.  197 

and  for  the  height  COi  =  hi  of  the  rolling  cone 


h.  =  r.  cotang  a.  =  r.  +  r.  cob  ja.  +  a,) 


sin  (jxx  +  aa) 


(7) 


Pi  J 


The  plane  through  the  instantaneous  axis  and  the  axis  of  the 
fixed  cone  passes  through  the  axis  of  the  rolling  cone  and  turns 
about  the  axis  of  the  fixed  cone  with  the  angular  velocity  ooi.  The 
motion  of  this  plane  is  called  the  precession  and  &Ja  is  the  angular 
velocity  of  the  precession,  or,  as  it  is  sometimes  called,  the  rate  of 
precession. 

If  a-i  is  zero,  the  fixed  cone  becomes  a  cylinder.  If  ai  is  zero,  the 
rolling  cone  becomes  a  cylinder.  If  both  are  zero,  we  have  the  case 
of  the  preceding  Article. 

If  oit  is  less  than  90"  and  ai  is  less  than  90°,  we  have  a  convex 
cone  rolling  on  a  convex  cone,  and 
looking  from  C  along  the  axes  COs 
and  CI  the  precession  and  rotation 
about  the  instantaneous  axis  are  both 
clockwise  or  both  counter-clockwise. 
This  is  called  positive  processional 
rotation.  It  is  the  case  of  a  pair  of 
bevel-gear  wheels,  or  of  a  spinning  top 
whose  point  is  at  rest. 

If  aa  is  a  right  angle,  the  fixed  cone 
becomes  a  flat  disk  with  centre  at  C.  If  ai  is  a  right  angle,  the  roll- 
ing cone  becomes  a  flat  disk  with  centre  at  C.    If  a^  is  a  right  angle 
and  aa  is  zero,  we  have  a  cylinder  rolling  on  a  plane. 

If  iXi  is  obtuse,  we  have  a  convex  cone 
rolling  inside  a  concave  one,  and  looking 
from  C  along  the  axes  COa  and  CI,  if  the 
precession  is  counter-clockwise  the  rotation 
about  CI  is  clockwise  or  vice  versa.  This  is 
negative  precession.  It  is  the  case  of  the  pro- 
cessional motion  of  the  earth's  axis. 

If  ai  is  obtuse,  the  rolling  cone  becomes 
concave  and  we  have  a  concave  cone  rolling 
on  a  convex  cone.  This  is  also  positive  pre- 
cession. 
The  path  described  by  a  point  relatively  to 
the  fixed  axis  is  called  a  spherical  epitrochoid 
or  hypotrochoid  according  as  the  rolling  cone 
is  outside  or  inside  of  the  fixed  cone.  The 
special  form  of  spherical  epitrochoid  or  •  l'r.-o.li 
hypotrochoid  described  by  a  point  in  the  sur- 
face of  the  rolling  cone  is  a  spherical  epicycloid 
or  hypocycloid. 

Analytical  Determination  of  Resultant  Angular  Velocity  and 
Velocity  of  Translation  for  a  Rigid  System  with  Any  Number  of 
Non-concurring  Angular  Velocities.*  —  (Compare  Statics  —  Non- 
concurring  Forces.)  Take  any  point  O  of  the  rigid  system  as 
the  origin  of  a  system  of  rectangular  co-ordinates.  Let  the  com- 
ponent angular  velocities  be  ca,,  c^-2,  etc.,  making  the  angles  («.,  /Si, 
rx\  (aa,  /ia,  r^),  etc.,  w-ith  the  axes  of  X,  F,  Z,  respectively. 

*  Angular  accelerations  are  treated  in  precisely  the  same  way  as  angular 
velocities,  and  every  equation  in  this  Article  can  have  oo  replaced  by  a. 
The  student  should  make  such  substitution  and  interpret  the  results. 


198  RIGID  SYSTEM — TRANSLATION  AND  KOTATION.    [CHAP.  IL, 

Resultant  Angular  Velocity. — We  have,  just  as  on  page  65,  re- 
placing V  by  cj,  for  the  component  angular  velocities  parallel  to  X, 
Y,  Z, 

eox  =  <»i  cos  ai  +  oOi  COS  a^  +  .  .  .  =  2ol)COS  a ;  1 

a>l,  =  ooi  COS  /ii  +  ooi  COS  pi  +  .  .  .  =.  20JC0S  /H;  J-  •  •     •     •     (1) 
GOz  =  ooi  COS  Xi  +  oOi  COS  /a  +  .  .  .   =  ^&>COS  y.  J 

The  resultant  angular  velocity  is 


ojr  =  |/ffi>a;*  +  aJj/"  +  t»/, .      (2) 

and  its  direction  cosines  are 

cosa= — ,    cose=-^,    cos/=-— (3; 

COt  OOt  GOf 

The  magnitude  and  direction  of  the  resultant  angular  velocity 
are  thus  determined. 

Resultant  Velocity  of  Translation. — Let  (Xi,  yi,  Zi),  {Xi,  y^,  Zj), 
etc.,  be  the  co-ordinates  of  points  on  the  rotation  axes  of  wi,  gljj.  etc. 
We  can  resolve  each  angular  velocity  oji,  cou,  etc.  (page  190j,  into  an 
equal  angular  velocity  about  a  parallel  axis  through  the  origin  O, 
and  a  velocity  of  translation  of  the  system  due  to  a  couple,  given  by 
the  moment  of  «»j,  <»2,  etc.,  with  reference  to  O.  We  can  thus  re- 
duce the  given  angular  velocities  to  a  resultant  angular  velocity  o^,- 
about  an  axis  through  the  origin  O,  whose  magnitude  and  direction 
are  given  by  (1),  (2)  and  (3),  and  a  velocity  of  translation  v,-  of  the 
axis  through  O.  The  components  Vx ,  Vy ,  Vz  of  this  velocity  of  the 
axis  through  O,  along  the  axes  of  X,  Y,  Z,  are  therefore  given  by 
(compare  Statics — Non-concurring  Forces) 

Mx  =Vx=  2a)y  cos  X  —  2a)Z  COS  /S ;    ^ 

My  =  Vy=  2a)Z  COS  a  —  2oox  COS  y;  > (4) 

Mz  =  Vz  =  2oox  COS  /3  —  2ooy  cos  a.  J 

For  any  other  point  P  whose  co-ordinates  are  {x',  y",  z'}  we  have 
simply  to  put  x  —  x',  y  —  y',  z  —  z'  in  place  of  x,  y,  z  in  (4)  and  we 
have  for  the  components  of  the  velocity  of  this  point  along  the 
axes 

Vx  =  2a}y  COS  y  —  2ooz  COS  /3  +  {ooyZ'  —  oozy') ;  ] 

Vy  —  2ooZ  COS  a  —  ^caa?  COS  y  +  {a)zX'  —  gJxZ')  ;    I  •        •      •      (5) 

Vz  =  SooXCOS  /3  —  2ooy  COS  a  -\-  {ooxy'  —  ooyX).  J 

Let  us  write 

Vx  =  a>yZ'  —  oozy' ;  1 

Vy  =  OOzX'  —  GOxZ' ;    l (6) 

Vz     =   OOxy'  —  GOyX'.    J 

Then  we  can  write  in  general  for  the  components  of  the  linear 
velocity  of  any  point  whose  co-ordinates  are  {x\  y',  z') 

Vx  =  Vx  +  Vx  ;  1 

Vy=Vy  +  Vy'\     1 (7) 

Vz=Vz  +  Vz;    J 

where  Vx,Vy^  Vz  are  given  by  (4)  and  Vx',  Vy',  Vz  by  (6). 


CHAP.  II.]  CONDITIOlf  FOR  A  SIlfGLE  AKGULAK  VELOCITY.  199 

If  the  resultant  axis  of  rotation  passes  through  the  origin  O  we 
have  vx  =  0  uj,  =  0,  vz  =  0.  Therefore  equations  (6)  give  the  com- 
ponents of  the  hnear  velocity  of  the  point  P  due  to  rotation  about 
an  axis  through  the  origin  O  paraUel  to  the  resultant  axis 

ihe  resultant  linear  velocity  for  any  point  is  then  in  general 

Vr=\/Vx'+Vy'+Vz'; (8) 

and  its  direction-cosines  are 

cosa=^,    co86  =  ^,    cosc  =  ^ (9) 

The  magnitude  and  direction  of  the  resultant  linear  velocity  of 
any  point  are  thus  determined. 

Conditions  of  Rest.*— If  the  system  is  at  rest  we  must  have,  Ist, 
Vx,  Vy,  Vz  equal  to  zero,  or,  from  (7),  Vx  =  0,  v„  =  0,  Uz  =  0;  and 
also,  2d,  v'x  =  0,  V  y  =  0,  v'z  =  0.  We  see  from  (6)  that  the  second 
condition  is  fulfilled  when  a)x=0,  coy  =  0,  ajz  =  0,  that  is,  when  oor 
=  0  or  there  is  no  rotation.  In  this  case  all  the  angular  velocities 
must  reduce  to  two  equal  and  opposite  resultant  angular  velocities. 
The  first  condition  is  fulfilled  when  equations  (4)  are  zero.  That  is, 
the  two  equal  and  opposite  resultant  angular  velocities  must  pass 
through  the  same  point,  so  that  their  moment  is  zero  at  any  point. 

We  have  then  for  the  equations  of  condition  of  rest,  from  (1), 

2aoe08a  =  0;  ] 

2&jCOS/S  =  0;   I (10) 

Saocosy  =  0;  J 
and,  from  (4) 

Saoy  cos  r  —  ^ooz  cos  /?  =  0;  "j 

SooZCOaa  —  Saoxcosy  =  0;   I (H) 

2oox  COS  /3  —  2ooy  cos  a  =  0.  J 

If  equations  (11)  only  are  fulfilled,  then  the  two  opposite  resultant 
angular  velocities  pass  through  the  origin,  which  is  therefore  at 
rest;  but  unless  (10)  is  also  fulfilled  they  are  not  equal,  and  we  have 
rotation  about  an  axis  through  O,  but  no  translation. 

If  equations  (10)  only  are  fulfilled,  there  is  no  rotation,  the  two 
resultant  angular  velocities  are  opposite  and  equal,  but  unless  (11) 
is  also  fulfilled  they  do  not  pass  through  the  same  point.  Hence 
they  form  a  couple,  and  we  have  translation  and  no  rotation. 

Condition  that  the  Angular  Velocities  shall  Reduce  to  a  Single 
Angular  Velocity. — If  the  angular  velocities,  then,  all  intersect  in 
one  point  of  the  system,  the  moment  at  that  point  is  zero.  It  has 
therefore  no  translation  and  the  system  rotates  about  an .  axis 
through  that  point.  If  the  angular  velocities  do  not  intersect  in  a 
single  point,  we  have  in  general  translation  and  rotation. 

There  is,  however,  one  case  in  which  the  angular  velocities  may 
not  all  intersect  in  one  point,  and  yet  we  may  have  rotation  only 
without  translation.  In  this  case  the  angular  velocities  must  re- 
duce to  three,  any  ttco  of  which  intersect,  while  the  other,  although 
it  does  notpdss  through  ilieir point  of  intersection,  yet  intersects  their 
resultant. 

*  Compare  Statics — Non-concurring  Forces. 


200 


RIGID  SYSTEM— ^TBANSLATIOK  AND  KOTATION.    [CHAP.  II. 


Thus  let  the  resultant  angular  velocities  gox,  coy  intersect  in  a 

point  A.  We  can  then  take 
them  as  acting  at  any  point  in 
their  resultant  AC. 

Let  ooz  intersect  AC  at  S. 
Then  we  can  take  all  three 
acting  at  JB,  and  we  thus  have 
rotation  only,  about  an  axis 
through_5^  _ 

Let  ar,  y,  z  be  the  co-ordi- 
nates of  B.  Then,  since  we 
can  take  oox,  a>y,  ooz  at  B,  we 
have  for  the  components  of 
the  velocity  of  the  origin  O     , 


(12)' 


My 

Mx  =  Vx  =  oozy  —  a>yZ ; 

^y  =  Vy  =   OOxZ  —   OOzX'i 
Mz  =  Vz  =^  OOyX  —  ooxy. 

If  we  multiply  the  first  of  these  by  oox ,  the  second  by  coy ,  and 
the  third  by  wz  and  add,  we  have  (compare  Statics — Non-concurring 
Forces) 

Vx  OOx  +  Vji  QOy  +  Vz  OOz  =^  0 (13) 

We  should  obtain  the  same  result  for  any  other  two  components 
intersecting  and  a  third  passing  through  a  point  on  their  resultant. 

Equation  (13j  then  gives  the  condition  that  all  the  angular  ve- 
locities acting  upon  the  system  reduce  to  a  single  angular  velocity 
at  a  point  whose  co-ordinates  are  x,  y  and  z,  and  we  have  rotation 
only  (page  179). 

We  have  evidently  for  the  equations  of  the  projection  of  the 
line  of  the  resultant  on  the  co-ordinate  planes 

GOu  Vz  OOx  Vv  GOz  Vx 

y  =  -!Lx  ~ ,        X  =  Z  —    ^-,        Z  ^  ^^y . 

<ax  GOx  OOz  GOz  OOy  OOy 

Parallel  Velocities.^ — (Compare  Statics — Non-concurring  Forces.) 
If  the  axes  of  all  the  angular  velocities  are  parallel,  we  have  «,  /^,  y 
constant  and  the  same  for  all.    Hence  from  (3)  and  (1) 

OOx  =  oor  COS  d  =  cos  a'Soo ;   1 

Wy  =  oor  COS  e  =  COS  /32a3 ;    [► (14) 

OOz  —  GOr  COS  /  =  COS  y2oo. 

The  resultant  oor  must  have  the  common  direction  of  the  paral- 
lel components,  or 

d  =  a,    e  —  (i,    f  =  r,    and    ca^  =  ^oo ,-....    (15) 

that  is,  the  resultant  angular  velocity  is  equal  to  the  algebraic  sum 
of  all  the  parallel  components  and  is  parallel  to  them. 

For  any  point  of  the  system  whose  co-ordinates  are  x',  y',  z',  we 
have  from  (4),  by  putting  {x  —  x'},  (y  —  y'),  (z  —  z')  in  place  of  x,  t/, 
z  and  taking  a,  (i,  y  as  constant, 

V^  —  cos  -{tioiy  —  y')  -  cos  fi'S.uii.z  -  z')  =  cos  7[2a>//  -  ?/'2(o]  -  cos  /SrSux  -  z'Soj]  ;    \ 

F„  =  cos  a2<u(Z  -  Z')-  COS  y^S.u,(x  -  x')  =  cos  a[2a>Z  -  z'2a>]  -  COS  ■)  [2wx-  -  a:'2<o] ;   y  .       (16^ 

Fg  =  COB  P2«(«—  »•)  -  COS  aXmiy  -  y')  =  COS  (3[2wa;  -  a:'2w]  -  cos  a[2w^-  j/'2w].  ) 


CHAP.  II.]   COMPONENTS  OF  MOTION  OF  A   RIGID  SYSTEM.  201 

If  we  substitute  (16)  and  (14)  in  (13),  we  see  that  equation  (13)  is. 
satisfied.  We  have  therefore  a  single  resultant  velocity  and  rota- 
tion of  the  system  about  a  fixed  point.  This  point  is  given  by  the 
values  of  x\  y',  z'  which  make  Vx,  Vy,  Vz  zero.  The  point  there- 
fore whose  co-ordinates  are 

—        2(ax     —        2coy      —        2a)Z 

*=^^'    2/=-^.    ^=-2^' (17) 

is  at  rest  and  the  resultant  axis  must  pass  through  it.  This  point 
is  called  the  centre  of  parallel  velocities  (page  193).  Any  other 
point  has  a  velocity  given  by  (16).  If  2 go  =  0,  the  resultant  axis  is 
at  an  infinite  distance,  or  there  is  no  rotation,  but  translation  only, 
given  by  (16). 

Components  of  Motion  of  a  Rigid  System.— In  order  that  the 
motion  of  a  rigid  system  at  any  instant  may  be  known,  it  is  sufli- 
cient  to  know  the  velocity  at  that  instant  of  some  given  point  of  the 
system,  and  the  rotation  of  the  system  at  that  instant. 

Take  the  given  point  always  as  the  origin.  Then  the  velocity  of 
this  point  is  known  when  its  components  Vx,  Vy,  Vz  along  the 
axes  are  given,  and  the  rotation  is  known  when  the  components 
Gox,  ooy,  Goz  of  the  augular  velocity  along  the  axes  are  given. 

The  motion  of  the  system  at  any  instant  is  then  known  when 
these  six  quantities,  Vx,  Vy,  Vz ,  cox,  ooy,  ooz  are  given.  These  six 
quantities  are  called  the  components  of  motion  of  the  system. 

Equivalent  Screw. — (Compare  Statics — Non-concurring  Forces.) 
The  motion  of  a  rigid  system  being  thus  known,  it  is  required  to 
find  the  screw  motion  to  which  it  is  equivalent.  That  is,  to  find 
the  central  axis,  the  linear  velocity  along  the  central  axis,  and  the 
angular  velocity  about  it. 

Since  Vx,  Vy,  Vz,  oox,  wy,  aoz  are  given,  we  have: 

(1)  The  angxilar  velocity  about  the  central  axis 

GOr  =  Vo^x'  +  oOy''  +  '^z. (1) 

(2)  The  direction  cosines  of  the  central  axis 

cos  d  =  -^,  cos  e  =  — ^,   cos/  =  — (2) 

ca,.  00^  oor 

(3)  The  linear  velocity  of  every  point  resolved  in  a  direction 
parallel  to  the  central  axis  must  be  the  same  and  equal  to  that 
along  the  central  axis.  Let  u,-  be  the  resultant  linear  velocity  of 
every  point  of  the  system  along  the  central  axis,  and  let  its  compo- 
nents along  the  co-ordinates  axes  be  Ux,  Uy,  Uz- 

Take  the  point  for  which  Vx,  Vy,  Vz  are  given,  as  the  origin, 
and  let  the  co-ordinates  of  any  point  of  the  central  axis  be  xf',  y", 
z".  Then  the  components  Vx,  Vy,  Vz  of  the  velocity  of  the  origin 
due  to  rotation  about  the  central  axis  are,  from  equations  (1),  page 
179, 

Vx=  oDztf"  —  f^yz"; 

Vy  =  ooxZ"  —  o^zX";     ■ (3) 

Vz  =  ooyx"  —  o^xV". 

We  have  then 

Vx=Ux+  Vx,       Vy  =  Uy  +  Vy,       Vz  =  Uz  +  Vz  i 

or 

llx=  Vx  —  Vx,      'iMl=Vy—Vy,      Uz=Vz-Vz..      .      .      (4). 


202  EIGID  SYSTEM — TRAlfSLATIOK  AND  ROTATIOlf.    [CHAf*.  II. 

Hence 

Mr  =  {Vx  —  Vx)  COS  d  +  iVy—  Vy)  COS  6  +  (Vz  —  Vz)  COS/.        (5) 

Inserting  the  values  of  the  direction-cosines  of  the  central  axis 
from  (2),  we  obtain 

Ur  a)r={Vx—  Vx)0Ox  +  (Vy  —  Vy )0Oy  +  (Vz  —  Vz)(^z. 

But  since  Vx^x  +  Vy^y  +  Vz  ooz  =  0,  this  becomes 

Vn-  oa,-  =   VxO!>X  +   Vy  (>^y  +    Vzl^Z (6) 

We  also  have  from  (4) 

Mr  COS  d=Ux=Vx—  Vxi    «*r  COS  6  =  Vy  —  Vy ,   Ur  COS  f  =  Vz  —  Vz.      (7) 

Hence  from  (2)  and  (3) 

Ur  Vx+  f^yZ''  —  oozy  '  _Vy  +  oOzX"  —  oo^z"        Vz  +  ^xV"  —  <^yX"     ._. 

=  ■ —  — .    ( o » 

cOr  oox  ooy  ooz 

Equations  (8)  give  the  equation  of  the  central  axis. 
From  1 6 )  and  (1)  we  have 

Ur^  _  Vx^X  +    VyCOy  +    Vz^Z  ^gx 

This  we  have  called  the  unit  pitch  (page  191),  or  the  distance  of 
advance  during  a  rotation  of  one  radian  about  the  central  axis. 

If  we  substitute  (9)  in  (8j  and  reduce,  we  have  for  the  equation 
of  the  central  axis 

1    /    ,,        VzOOy  —  Vyooz  \  _    1    /    ,,         VxOOz  —  VzGOx  \ 
oox  \  <»r'  l~  <^y\  <»'•'  I 

.       ^±L._Vy^x-VxC^y\        ^     (^0^ 


Therefore  the  central  axis  passes  through  a  point  whose  co- 
ordinates are  * 

_  FgCJy  —  VyC^z  „  _  Vxf^z  —  VzCOx  »  _  VyOOx  —  VxOOy     .^^. 

If  we  substitute  these  values  of  x",  y",  z",  in  (3)  and  (7),  we 
have  from  (2) 

Vx  =  u,-  cos  d  —  <»r(2;"  COS  e  —  yi'  cos/),   <»«  =  <»>•  cos  d\  1 

Vy  =  Ur   COS  6  —  oa,(a^'  COS  /  —  z"   COS  d),     OOy  =  OOr   COS  6;  |-  .   (12) 
Vz^'Ur   COS  /  —  ^r(y"   COS  d  —  ic"  COS  e),  <»2  =  ca,-  COS/. 

When,  therefore,  the  components  of  motion  Vx,  Vy,  Vz,  f^x, 
coy,  ooz  are  given  for  any  point,  we  find  <»,  from  (1),  the  direction 
of  the  central  axis  from  (2),  and  the  position  of  the  central  axis 
with  reference  to  that  point  as  an  origin  from  (11).  We  have 
also  the  velocity  of  advance  Ur  from  (9). 

*  Since  velocity  in  the  liodograph  is  normal  acceleration  in  the  path  (page 

52),  Vzooy  —  Vyooz  is  the  component  in  the  direction  of  Xoi  the  normal  linear 

acceleration  of  the  origin  due  to  rotation  about  the  central  axis.     The  normal 

Vzo^v  —  VyOOz 
linear  acceleration  is  poor^.     Hence — ^— ^ is  the  projection  of  p  on  the 

«xis  of  X. 


CHAP.  II.]      COMPOSITION  AND   RESOLUTION   OF   SCEBWS.  203 

On  the  other  hand,  if  the  position  of  the  central  axis  'x"  v"  z") 
IS  known  together  with  the  linear  velocity  Ur  along  it  and  the 'an- 
gular velocity  oor  round  it,  the  components  of  the  motion  for  the 
origin  are  given  by  (12). 

The  Invariant.— (Compare  Statics— Non-concurring  Forces) 
From  (6)  we  see  that  the  quantity 

VxOfx  +  VyGi)j/  +  VzGOz 

is  always  equal  to  m,  to, ,  and  is  therefore  invariable  no  matter  what 
point  is  taken  and  whatever  the  values  of  a>x,  ooy,  aoz,  that  is,  what- 
ever the  direction  of  the  axes.  This  quantity  is  therefore' called 
the  invariant  of  the  components.  Since  oo,.  is  also  invariable  what- 
ever point  is  taken  and  whatever  the  direction  of  the  axes,  it  may 
be  called  the  invariant  of  the  rotation. 

If  the  motion  is  such  that  the  invariant  is  zero,  it  follows  that 
either  m,-  =  0  or  ca,-  =  0.     The  condition 

VxOOx  +    VyCOy  +    VzOOz  =  0 

is  therefore  the  condition  that  the  motion  is  equivalent  to  either  a 
simple  translation  or  a  simple  rotation.  If  &?)•  is  not  zero  and  this 
condition  is  fulfilled,  we  have  rotation  only  (see  pages  179,  200). 

Composition  and  Resolution  of  Screws.  —  (Compare  Statics— 
Non-concurring  Forces.)  If  two  screws  are  given,  then  by  equa- 
tion f  12)  we  can  find  the  six  components  of  motion  of  each  screw. 
Adding  these  two  and  two,  we  have  the  six  components  of  the 
resultant  screw.  Then  by  (1),  (2),  (6)  and  (11)  the  central  axis 
together  with  the  linear  and  angular  velocities  of  the  screw  may 
be  lound. 

Conversely,  we  may  resolve  any  given  screw  motion  into  two 
screws  in  an  infinite  number  of  ways.  Since  a  screw  motion  is 
represented  by  six  components  at  any  point,  we  have  in  the  two 
screws  twelve  quantities  at  our  disposal.  Six  of  these  are  required 
to  make  the  two  screws  equivalent  to  the  given  screw.  We  may 
therefore  in  general  satisfy  six  other  conditions  at  pleasure. 

Thus  we  may  choose  the  axis  of  one  screw  to  be  any  given  straight 
line  we  please  with  any  given  linear  velocity  along  it  and  any  angu- 
lar velocity  round  it.  The  other  screw  may  then  be  found  by  re- 
versing this  assumed  screw  and  joining  it  thus  changed  to  the  given 
motion.  The  screw  equivalent  to  this  compound  motion  is  the 
second  screw,  and  it  may  be  found  in  the  inanner  just  explained. 

Or  again,  we  may  represent  the  motion  by  two  screws  whose 
unit  pitches  are  both  zero,  the  axis  of  one  being  arbitrary.  We  can 
thus  represent  any  motion  by  two  angular  velocities,  one,  <»,  about 
an  axis  which  we  may  choose  at  pleasure,  and  the  other,  w',  about 
some  axis  which  does  not  in  general  cut  the  first  axis.  These  are 
called  conjugate  axes. 

These  angular  velocities  are  such  that  air  would  be  their  resultant- 
if  their  axes  were  placed  parallel  to  their  actual  posi-     ^ 
tions,  so  as  to  intersect  the  central  axis.    If  then  d  is  the      ^\i^ 
shortest  distance  between  the  axes,  we  have  Fr  =  dcj; 
and  if  ^  is  the  angle  between  &?  and  «r ,  and  0  the  angle 
between  co  and  oo,  we  have 


Also 


Ur 

Vr  sin  ^  =  ttr ,  or    smip  =  ==-. 

Vr 

.    .  .     ,       fij'  sin  9 

fijr  sin  if}  =  00  sm  9,    or    sin  ip  = 

cor 


204  RIGID  SYSTEM — TRANSLATION  AND  ROTATION.     [CHAP.  11^ 

Hence 

"Fi-fij'sine  =tlr00r,     or     doooo'ainB  =  Uroor', 

hence 

UrCOr 


d  = 


Gooa'  sin  9 


(1> 


EXAMPLES. 


(1)  A  line  DE  moves,  keeping  its  extremities  in  tico  fixed  lines 
ADB,  AEC.  Find  the  instantaneous  centre  and  the  direction  of 
motion  of  any  point  G  at  any  instant. 

Ans.  From  D  and  E  draw  BF  and  EF  perpendicular  to  AB  and  A  C,  meet- 
ing at  P.  Then  F'  is  the  instantaneous  centre  (page  189).  Join  OF.  The  di- 
rection of  motion  of  G  is  perpendicular  to  OF. 

(2)  A  line  DE  moves,  keeping  its  extremities  in  two  fixed  lines, 
one,  ADB,  vertical  and  the  other,  AEC,  horizontal,  and  makes  at  a 
given  instant  an  angle  of  30°  with  the  horizontal.  Find  (a)  the  di- 
rection of  motion  of  the  middle  point  of  DE  at  the  instant,  and 
(b)  the  point  whose  motion  is  inclined  at  that  instant  30°  to  AC. 

Ans.  (a)  Inclined  60°  to  ^C ;  (b)  ~DE  from  E. 

(3)  A  line  moves  so  that  its  extremities  remain  in  a  given  circle. 
Find  the  instantaneous  centre  of  motion  for  any  instant. 

Ans.  The  centre  of  the  circle. 

(4)  Find  the  ratio  of  the  velocity  of  any  point  of  a  screw  to  its 
velocity  of  advance. 


/ 


d 


where  d  is  the  pitch,  r  the  radius  of  the  screw  (page 


Ans, 
191). 

(5)  Let  Ci  and  d  he  fixed  axes  about  which  turn  the  cranks  CA\, 
CiB,  whose  free  ends  are  connected  by  the  link  AB,  jointed  at  A  and 
B.  The  axes  are  perpendicular  and  the  plane  of  motion  parallel  ta 
the  paper.  If  the  linear  and  angidar  velocities  of  A  are  Vi,  ooi, 
find  the  linear  and  angular  velocities  Vt ,  wi  of  B. 

Ans.  Let  ACi=ri  and  BCt  —  r^.  Produce  CiA  and  dB  to  meet  in  J. 
Then  at  any  instant  the  linear  velocities  of  A  and  B  are  perpendicular  to  A  C» 


and  BGi  respectively.  Hence  at  that  instant  AB  is  rotating  about  the  instan- 
taneous axis  at  /.  Let  co  be  the  angular  velocity  of  AB  about  /.  From  d  , 
Ca,  /let  fall  the  perpendiculars  C\D,  VtF,  IE  on  the  line  of  the  link  AB  or 


CHAP.  II.] 


EXAMPLES. 


205 


its  prolongation.     Also  draw  the  line  of  centres   Ci  C,  cutting  the  link,  pro- 
longed, if  necessary,  in  the  point  K.     Then 


TidJi  =  AI .  00,     or     —  = 


rjftJa  =  BI.  CO,     or 


also,     — 


Al_ 

IE 
CD' 

BI  _ 

IE 

C,F 

C,K 

Since  naji  =  Vi  and  r^oo,  =  tJa ,  we  have 
Ti  _BI 

v,~  AI  _ 

Hence — 

1.  The  linear  velocities  of  B  and  A  are  to  each  other  as  their  distances  from 
the  instantaneous  axis. 

2.  The  angular  velocities  of  the  cranks  are  to  each  other  inversely  as  the 
perpendiculars  from  their  centres  of  motion  upon  the  line  of  the  link  ;  or  in- 
versely as  the  segments  into  which  the  line  of  the  link  cuts  the  line  of 
centres. 


(6)  In  the  case  of  the  crank  and  connecting  rod,  since  B  moves  in  a  straight 


line  CiB,  we  have  BI  always  perpen- 
<licular  to  GiB,  and  hence 


«i 


BI 

AI' 


or    Vi  = 


AI 


TiGOi. 


Let  the  distance  CiB  =  s,  the  length 
of  the  connecting-rod  =  I,  and  the  angle 
of  the  crank  r,  with  CiB  =  61.  Then 
we  have 


BI=  8  tan  I 


AI  = 


cos  01 


-  n, 


s  =  ri  cos  6,  +  Vi*  -  ri^  sin*  6,; 
or,  if  I  is  very  long  compared  to  ri ,  approximately 

ri*  sin*  fli 


>.t?. 


8  =  Ti  cos  61  -\-  I  — 


21 


Hence 


«j  = 


Stan 01  .  TiOOi 


«  sin  01 


cos  01 


—  Ti 


8  —  r,  cos  Bi 


-rifiJi. 


When  01  =  90,  we  have  v,  =  rjcoi  =  «i ,  or  the  velocity  of  ^  and  5  are 
equal  and  B/and  Alure  infinite.  When  0i  =  0  or  180°.  we  have  e,  =  0  and 
s  =  l-\-ri  or  I—  r,.  These  are  the  "  dead  points  "  of  the  crank,  or  the  ends 
of  the  stroke. 

(7)  A  rod  (length  -  I)  hangs  by  a  small  ring  at  its  upper  end 
from  a  fixed  horizontal  rod.  To  the  former  an  angular  velocity  go 
is  given  in  a  vertical  plane  through  th^  fixed  rod,  so  that  the  centre 
of  the  movable  rod  moves  vertically.  Find  the  linear  velocity  of 
its  centre  when  its  inclination  to  the  vesical  is  0. 

Ans.  GolsinO. 

(8)  A  disk  (radius  =  r)  rolls  without  sliding  on  a  plane.  Find 
the  relation  between  its  angular  velocity  <»  and  the  linear  velocity  v 
of  its  c&Yvtv^t 

Ans.  The  point  of  contact  with  the  plane  is  at  rest  at  any  instant,  or 
ra>  =  —  ». 


206 


RIGID  SYSTEM — TRANSLATION  AND  ROTATION.    [CHAP.  II. 


(9)  A  rod  AB  (length  =  J)  rotates  about  a  hinge  at  A  and  rests 
with  its  end  B  on  the  surface  of  a  ivedge  BCD.     The  wedge  advances 

towards  A  xvith  velocity  v.  The  angles 
BAG  =  6.  BCD  =  (p.  Find  the  angular 
velocity  oo  of  the  rod. 

V  sin  <p 


Ans. 


I    cos  (0  —  0)' 

flO)  Two  bevel-gear  wheels  have  the  angle  between  their  axes  70°. 
The  rolling  wheel  is  required  to  make  3i  revolutions  about  its  axis 
while  going  around  the  axis  of  the  fixed  ivheel  once.  Find  the  angles 
of  the  bevel.  If  the  inner  radius  of  the  fixed  ivheel  is  50  inches,  find 
that  of  the  rolling  wheel  and  the  length  of  the  axes. 

Ans.  (See  page  196.)     The  angular-velocity  ratio 
<Bl        7        „ 

—  =  - .     Hence 

GO,        2 

7  sin  70°  _-o  . ^, 

tang  or,  =  »  ,  r,„„„;^o,    or    a,  =  56   15', 


and  Lence   a, 


2  +  7  cos  70 
13°  45'. 


inches,    and  hi  = 
100 

7 


We  have  also  ri  =  -=- 

^  -I-  50  cos  70° 

=  33.4   inches, 


sin  70 


50  + 


cos70° 


A. 


sin  70° 


=  58.4  inches. 


(11)  The  angle  between  the  plane  of  the  earth''s  equator  and  the 
plane  of  the  ecliptic  is  23°  27'  28".  The  earth  rotates  about  its  polar 
axis  in  one  sidereal  day  and  makes  a  revolution  about  the  axis  per- 
pendicular to  the  plane  of  the  ecliptic  in  25868  years.  Finathe 
instantaneous  axis. 

Ans.  (Page  196.)     We  have 

fij,  =  27r  radians  per  day,     and    co^  =    ^„  „ — g^rr  radians  per  day. 


Also  the  angle  O^COi  =  ai-{-  a 

Oil  =  r  tan  ai 


25868  X  865i 
23°  27' 28".     Therefore  0,/ is 
r  sin  (flTi  +  aa) 


OOi 


-\-  cos  (ai  -f  aj) 


where  r  is  the  polar  radius  of  the  earth,  or  3950  miles.   The  radius  of  the  roll- 
ing cone  is  then  Oil—  5.52  ft.  and  the  angle  a,  =  0". 00867. 

(13)  A  rigid  system  has  an  angular  velocity  of  ooi  =  40  radians 
per  sec.  about  an  axis  parallel  to  the  axis  of  X,  passing  through  a 
point  whose  co-ordinates  are  Xi  =  2  ft.,  yi  =  3ft.,  Zi  =  0.  and  a 
simultaneous  angular  velocity  of  oo^  =  30  radians  per  sec.  about  an 
axis  coinciding  with  the  axis  of  Y.  Find  the  resultant  angular 
velocity  and  the  instantaneous  axis. 

Ans.  (Page  198.)  We  have  ftjj- = -f- 40,  ajj/zr-j-SO,  ea^  =  0,  <»,  =  50  ra- 
dians per  sec.  The  instantaneous  axis  makes  the  angles  with  the  axes  given 
by 


cos  o  ^  — , 

50 


30 
cose  =  ~,      cos/: 


0; 


d  =  36°  52',    e  =  53°  7',      /=  0°. 


CHAP.  II.] 


EXAMPLES. 


207 


Uz 


•*-GOx 


The  velocity  of  translation  of  the  system  is  given  by 
ux  =0,    uy  =  0,    uz  =  —  120  ft.  per  sec.  a)|] 

The  condition  uxoox  +  Uyooy  -\-  uzooz  =  0  is  sat- 
isfied. Therefore  there  is  rotation  only  about 
the  instantaneous  axis  which  passes  through  the 
intersection  /of  oox  and  cay. 

The  velocity  of  any  point  whose  co-ordinates 
are  x  =  2,  i/'  =  2,  2'  =  3  is  given  by 

Vx  —  ooyz'  =  -|-  90  ft.  per  sec. ; 

Yy=  —  oox!^  —  —  120  ft.  per  sec. ; 

Vz  =  Uz  —  ooyx'  4-  o^a^  =  —  100  ft.  per  sec. 

The  resultant  velocity  of  this  point  is  then  Vr  =  180  ft.  per  sec.  and  its 
direction  cosines  are 


90  ^      120  100 

cos  a  =  rr^x.        COS  0  =  tt^^,       COS  C  = 


,-      120 

cos  0  :=  . 

180' 


180'  "      180'     ^"""180' 

or         a  =  60°,  6  =  131°  48',    c  =  123°  45'. 

(13)  A  rigid  system  has  the  angular  velocities 
ttj,  =  50,     ttJj  =  30,     (»3  =  70,     cot  =  90,    and    ooi  =  120  radians  per  sec. 
about  axes  passing  through  points  of  the  rigid  system  given  by 
Xi=-h   5  ft.,  2^1  =+10  ft.;    «,=+    9  ft.,  2^j  =  +12ft,; 
a;3  =  +  17ft.,2/3  =  +  14ft.;    ««  =  +  20  ft.,  y*  =  + 13  ft. ; 
a;^  = -f  15  ft.,  ys  =  +   8  ft.; 
and  making  with  the  co-ordinate  axes  the  angles 

a.  =    70°./?!=    20°;     a,  =  60°, /?,  =  150°;     or,  =  120°, /Jj  =  0°; 
a,  =  150°,  /34  =  120°;     a,  =  90°,  /S.  =     0°. 
Find  the  resultant,  etc.     (Compare  Vol.  II,  Statics.) 

Ans.  (Page  198.)    We  have  for  the  components  of  the  angular  velocities 
parallel  to  the  axes 
cox  =  50  cos  70°  +  30  cos  60°  —    70  cos  60°  —  90  cos  30°  =  —  80.842  rad.  per  sec. ; 

ooy  =  50  cos  20°  -  30  cos  30°  + 120  +  70  cos  30°  -  90  cos  60°  =  + 156.626  rad. 

[per  sec. ; 
coz  =  0. 

The  resultant  angular  velocity  is  given  in  magnitude  by 

oor  =  '^oox*  +  ooy^  =  +  176.259  radians  per  sec, 
and  its  direction-cosines  by 


cox       -  80.843 

cos  d  =  = 

oor 


OOy 


176.259  ' 
+ 156.626 


or     <2  =  117°  18' 1"; 
or      e=    27°18'1". 


""°  "  ~  Or  176.259  ' 

We  have  from  equation  (4),  page  198, 


per  sec. ; 
,  per  sec 


"208  RIGID  SYSTEM— TRANSLATION  AKD  ROTATION.    [CHAP.  II. 

We  have  then  for  the  components  of  the  linear  velocity  of  the  origin 

Vx  =  0,     vy  =  0,     Vz  =  2 oox  cos /3  —  2a)ycx>s  a  = -\-d083.dl5  ft. -per  sec. 

Since  then  equation  (13),  page  200, 

VxOOx  -f-  i>y'*'y  -\-  '>^z^z  =  0 

is  satisfied,  the  angular  velocities  reduce  to  a  single  resultant  angular  velocity 
-and  we  have  rotation  only. 

The  moment  of  this  resultant  angular  velocity  relative  to  the  origin  gives  us 


«}•  =  \/vx'  -\-  %*  +  Bz*  =  «2  =  +  3083.915  ft.  per  sec. 

Its  lever-arm  is 

vr        3083.915       ,„^^^ 
r  =  -  =  ,-„  q.„  =  17.5  ft. 
oor       176. 3o9 

The  equation  of  the  line  of  direction  of  the  resultant  angular  velocity  is 
y  =  ^x  -  ^  =  -  1.95a;  +  38.14 

OOx  COx 

The  co-ordinates  of  the  point  through  which  this  resultant  angular  velocity 
passes  are  given  from  equations  (17),  page  201: 

X  =  ?^^2iA  =  +  I2i  ft.;    ^  =  2^yco.a  ^  ^  ^^^^^ 

COy  OOx 

(14)  Find  the  resultant  angular  velocity  for  a  number  of  parallel 
■angular  velocities  given  by 

ooi  =  -\-    83  rad.  per  sec;     «!  =  +  25  ft.;     jti  =  -f  13  ft.; 

(»!,  =  +    20    "      "     "         X2  =  —  10  "       Pi  =  —  15  " 

G03  =  -    35    "      "     "         X,  =  -f  15  "       y»  =  -  37  " 

(B4  =  -    72    "      "     "        0:4  =  -  31  '•       y4  =  +  17  " 

ttjs  =  -f  120    "      "     "         ais  =  -f  23  "       Vi  =  —  19  " 

Ans.  ttJr  =  -|-  66  radians  per  sec;  x  —  -[-  77.15  ft.;    ^  =  —  36.82  ft. 

(15)  Find  the  resultant,  etc.,  for  the  angular  velocities  given  by 

GOi  —  50,  002  =  70,  oiJs  =  90,  (»4  =  120  radians  per  sec. 


«,  =  60° 
a,  =  65° 
«,  =  70° 
at  =  75° 


/Si  =  40°;     Xi  acute;     Xi  =0;  yi  =  0;  a,  =  0; 

/3j  =  45°;     Xi  acute;    Xa  =  +  1  ft. ;    yj  =  -|-  4  ft. ;    «»  =  +  7  ft. ; 
/Sa  =  50":     rs  acute;    Xs  =  +2  "       2/3  =  -f  5  "       Zs  =  +  8  " 
/J4  =  55°;    ^^4  obtuse;  X4  = +8  "       ^4  =  +  6  "       a4  =  -j-9" 
(Compare  Vol.  II,  Statics.) 

Ans.  We  find  the  angles  y  by  the  formula,  page  12, 

cos*  y  =  —  cos  (a  -|-  /3)  cos  (a  —  /3). 
Then  from  page  198  we  have 

fiJx  =  -f  116.423,     ooy=  -{-  214.480,    ooz  =  -  51.057  rad.  per  sec. 
Therefore  the  resultant  angular  velocity  is 


oor  =  i/oox^  +  fi>i/*  -|-  02*  =  +  349.325  rad.  per  sec, 
and  its  direction-cosines  are  given  by 

cos  a  = ,     cos  e  = ,     cos/  = ; 

GO)-  oor  G^r 

or 

a  -  63°  9'  48",    e  =  30°  39'  20",    /  =  101°  49'. 


CHAP.  II.]  EXAMPLES.  209 

We  also  Lave  for  the  components  of  the  velocity  of  the  origin,  from  equa- 
tions (4),  page  198, 

vx=  -  1838.604,     %  =  -f  928.947.    Tz  =  -  86.903  ft.  per  sec. 
The  resultant  linear  velocity  of  the  origin  is  then 


vr  =  4/cx*  +  V  +  »«*  =  +  2061.789  ft.  per  sec., 
and  its  direction-cosines  are  given  by 

Vx  ,  Vy  Vz 

cos  a  =  — ,     cos  0  =  -=-,    cos  c  =  — ; 

Vr  Vr  Vr 

at 

a  =  153°  5'  40",    6  =  63°  14'  15",    c  =  92°  24'  56". 

The  equations  of  the  projections  of  the  resultant  angular  velocity  on  the 
co-ordinate  planes  are 

y  =  1.885a;  -f  0.746,    x  =  -  2.28z  +  18.19,     s  =  -  0.238y  -  8.57. 

We  see  that  vxoox-\-i}yooy-\-  Vzooz  does  not,  in  this  case,  equal  zero.  Hence 
^page  179)  the  angular  velocities  do  not  reduce  to  a  single  resultant,  but  to  a 
resultant  angular  velocity  about  the  central  axis  and  translation  along  that 
axis  due  to  an  angular- velocity  couple. 

The  resultant  angular  velocity  about  the  central  axis  is,  as  already  found, 
cor=^-{-  249.325  rad.  per  sec,  and  its  angles  d,  e, /with  the  axes  are  already 
found. 

The  co-ordinates  of  the  central  axis  are  given  by  equations  (11),  page  202 : 

^,  ^  OayVz-OO^y    ^  ^^g  ^    OO^x  -  OOxVz    ^  _^  j  g^g  ^^  , 

aof*  '  oor 

a"  =  '^^y-'^y^^  =  .  8.08  ft. 
car* 

The  resultant  linear  velocity  Ur  along  the  central  axis  is  given  by  equation 
(6),  page  202: 

GOr 
Its  direction-cosines  are  the  same  as  for  oor.    The  components  of  Ur  are 
given  by  equations  (7),  page  202  : 

ux  =  Ur  cos  d  —  —  19.481,     %  =  «r  cos  e  =  —  35,806, 
Uz  —  Ur  cos/  =  -f-  8.5238  ft.  per  sec. 

(16)  In  the  preceding  example  find  ivhat  the  co-ordinates  ic* ,  «« , 
z*  of  the  angular  velocity  oot  must  be  in  order  that  all  the  angular 
velocities  may  reduce  to  a  single  resultant  angular  velocity.  (Com- 
pare Vol.  II,  Statics.) 

Ans.  We  must  evidently  have  oox,  ooy ,  aoz ,  oor  and  the  angles  d,  e,  f  un- 
changed, since  changing  the  co-ordinates  x^,  y^,  z*  without  changing  the  mag- 
nitude or  direction  of  oo^  has  no  effect  on  the  magnitade  or  direction  of  the 
resultant  oor. 

We  have  then 

vx=-  659.571  -  93.262^4  -  68.829s4;  ) 

«j,=  -f  369.629  +  31.05924  +93.262*4;  )• (1) 

vz=  -  107.036  +  68.82fti!«  -  31.0592^4.  ) 

"We  have  as  the  equation  of  condition  for  a  single  resultant 

VxOOx  +  VyOOy  +  VzODz  =  0, 

or 

116.423Ba;  +  214.48PJ/  -  51.057*2  =  0. 

■Pa;  +  1.842ci,-0.4386oz=  0 (2) 


210  EIGID  SYSTEM— TRANSLATIOlf  AND  ROTATIOlf.    [CHAP.  !!► 

From  (1)  we  obtain 

(Vx  +  659.571)  31.059  +  {vy  -  369.639)  68.829  =  (b2  +  107.036)  93.262, 
or 

®a;+ 2.216% -3.003»z  =  + 481.034. (3> 

From  (2)  and  (3)  we  obtain 

0.374%  -  2.564«z  =  +  481.034. 

If  we  retain  for  vy  its  value  in  the  preceding  example,  -}-  928.947  ft.  per 
sec,  we  shall  have 

vz=:-  52.108,     'Ox=-  1733.975  ft.  per  sec. 

If  we  substitute  these  values  in  (1),  we  obtain 

93.262^4  +  68.829S4  =  +  1074.4; 

31.05924  4-  93.262a;4  =  +    559.308; 

68.829a!4  -  31.059^4  =  +      54.934. 
Hence 

a!4  =  -  0.333S4  +  5.997,    y*  =  -  0.738z4  +  11.520. 
If  then  we  assume  z^  =  0,  we  have  x*  =  -\-  5.997,  i/t  =  -}-  11.52  ft. 

(17)  Using  the  values  of  the  preceding  example,  find  the  point 
through  which  the  resultant  angular  velocity  passes.  (Compare' 
Vol.  II,  Statics.) 

Ans.  We  have  Ox  =  +  116.433,     coy  =  +  214.480,     coz  =  -  51.057,  oor  = 
+  249.325  radians  per  sec. ;    d=  62°  9'  48",  e  =  30°  39'  20 ',  /  =  101°  49' ; 
i>x=-  1733.975,  %  =  +  928.947,  vz  =  -  53.108,  Vr  =  +  1967.833  ft.  per  sec. ; 
a  =  151°  47',    b  =  61°  49'  53",    c  =  91°  31'  3". 

The  co-ordinates  x,  y,  z  are  given  (page  300)  by 

-  1733.975  =  002^  -  GJijZ  =  -    51.057y  -  214.480sl 
4-    928.947  =  co:^  -  oo^  =  +  116.423s  -f    51.057«; 

-  52.108  ^ooyX-a}j^=  +  214.480»  -  116.423^. 
Hence  we  obtain 

x=  -  2.2802s +  18.194; 

y  =  -4.2008s +  33.961. 

If  we  assume  i  =  0,  we  have  «  =  +  18.194,  p  =  +  33.961  ft. 
If  we  should  introduce  then  a  fifth  angular  velocity,  0!J5=  +  249.325,  whose 
direction  makes  with  the  axes  the  angles 

as  =  117°  50'  12",     /Ji  =  149°  20'  40",     y^  =  78°  11', 

passing  through  a  point  whose  co-ordinates  are  ajj  =  +  18.194,  j'6  =  -f  33.961 
and  Zi  =  0,  the  conditions  for  rest  (page  199)  would  be  satisfied,  and  we  should 
have  QOr  =  0,  Vr  =  0. 

(18)  A  point  of  a  rigid  system  rotates  about  an  axis  at  a  distance 
of  5  feet.  The  linear  displacement  of  the  point  is  8  ft.  Find  the 
angular  displacement  ana  the  direction  of  the  linear  displacement. 

Ans.  sin—  =  —  —  ~-    5  =  angular  displacement  =  106'  14'  =  1.858  ra- 
2        2r       5 

dians.     The  linear  displacement  makes  an  angle  of  36°  53'  with  the  radius  of 

rotation. 


-V,-+3t 


CHAP.  II.]  EXAMPLES.  211 

(19)  A  point  of  a  rigid  system  has  at  any  instant  the  component 
linear  velocities  Vx=  +Q,Vy=  -  18,  Vz  =  +  40  ft.  per  sec.,  and  the 
system  at  the  same  instant  rotates  about  an  axis  perpendicular  to 
the  plarte  of  XY  with  an  angular  velocity  of  6  radians  per  sec.  in  the 
direction  from  X  towards  Y.  Find  the  equivalent  screw  motion  of 
the  system.  '' 

Alls.  (Page  201.)  We  take  the  given  point  as  the  origin.  Since  the  axis  is 
perpendicular  to  the  plane  of  XY,  we  have  eaa;  =  0,  fij»  =  0,  02  =  (Br  =  +  6 
radians  per  sec. 

Since  the  condition   Fa;<»a;-}-  VyGOy-\-  Vzwz  =0  is  not  fulfilled,  we  have 
rotation  and  translation  combined,  or  screw 
motion. 

From  equation  (2)  we  have  cos  d  =  0, 
cose  =  0,  cos/=l,  or  the  central  axis  is 
parallel  to  the  axis  of  Z. 

The  position  of  the  central  axis  is  from 
equation  (11)  given  by 

i:"  =  -h3ft.,     y"  =  +  lft.,     2"=0. 

It  is  therefore  at  /as  shown  in  the  figure     ^ 
with  respect  to  the  given  point  0. 

Substituting  these  values  in  equation  (3)  ^i 

we  have  for  the  components  along  the  co-ordinate  axes  of  the  velocity  of  0  due 
to  rotation  about  the  central  axis 

««  =  +  6,     %  =  —  18  ft.  per  sec.,     Vz  =  0. 

Therefore  from  (4)  we  have  for  the  components  along  the  axes  of  the  trans- 
lation of  0, 

ux  =  0,    uy  =  0,    uz  =  +  40  ft.  per  sec. 

The  system,  therefore,  at  the  instant  in  question  rotates  about  an  axis 
through  /perpendicular  to  the  plane  of  XY,  in  a  direction  from  X  towards  Y, 
with  the  angular  velocity  of  6  radians  per  sec. ,  and  at  the  same  time  moves 
along  this  axis  in  the  direction  OZ  with  a  velocity  of  translation  of  40  ft.  per 
sec. 

If  the  axis  of  rotation  and  angular  velocity  do  not  change  in  direction  or 
magnitude,  the  system  advances  along  the  central  axis  during  a  rotation  of  one 
radian,  a  distance  equal  to  the  unit  pitch,  given  by  equation  (9),  viz.,  6J  ft.  In 
one  complete  rotation  then  it  advances  a  distance  of  27r  X  6f  =  20.9  ft.  This 
is  the  fitch  of  the  screw  motion  (page  191). 

The  velocity  at  any  point,  as  Pi  or  Pa ,  due  to  rotation  about  the  central  axis 
is  equal  to  /Pi  .  ooz  or  IP^  .  ooz  ,  where  /P  or  /Pa  is  the  radius  vector  or  per- 
pendicular from  the  point  upon  the  central  axis.  If  then  we  take  Pi  as  origin 
and  the  co-ordinates  of  /  are  a;  =  -f-  2  f t. ,  y  =  -|-  5  ft. ,  z  =  +  3  f t. ,  we  have 
from  (3),  for  the  components  of  the  velocity  of  Pi  due  to  rotation  about  the 
central  axis, 

©a;  = -|- 30  ft.  per  sec.,     ©j/ =  — 12  ft.  per  sec,     Vz  —  O; 

and  since  ««  =  0,  «j/  =  0,  ««  =  +40  ft.  per  sec,  the  components  of  the  total 
velocity  of  Pi  are,  from  (4), 

Fx  =  +  30  ft.  per  sec,     Fj/  =  -  12  ft.  per  sec,     Fa  =  +  40  ft.  per  sec 

In  the  same  way  for  the  point  Ps  ,  if  we  take  it  as  origin  and  the  co-ordi- 
nates of  /are  a;  =  0,  2/  =  +  5  ft.,  s  =  0,  we  have 

Da;  = +30  ft.  per  sec,     %  =  0,     Vz  =  0; 

Fc  =  +  30  ft.  per  sec,   Vy  =  0,   Fa  =  -f  40  ft.  per  seci 

If  the  velocities  Fc  and  Fs  of  the  point  Pa  do  not  change  in  direction  or 
magnitude,  we  have  the  case  of  a  system  translated  in  the  direction  0?  and 
rotating  about  an  axis  through  Pa ,  while  at  the  same  time  this  axis  has  a  ve- 
locity of  translation  in  a  straight  line  (page  193). 


212 


BIGID  SYSTEM — TRAIfSLATION  AND  ROTATION.    [CHAP.  II. 


The  motion  of  the  system  would  then  be  the  same  as  if  a  cylindrical  surface 
of  radius  P.il=  5  ft.,  fixed  to  the  system  with  its  axis  passing  through  Pj  at 
right  angles  to  the  plane  of  XT,  rolled  on  the  plane  HIH  parallel  to  the  plane 
Xywith  the  angular  velocity  goz  —  -^Q  radians  per  sec,  while  at- the  same 
time  the  cylinder  is  translated  parallel  to  OZ  with  the  velocity  Fz  =  -(-  40  ft. 
per  sec. 

(20)  A  base-ball  rotates  about  an  axis  through  its  centre  in  a 
horizontal  plane  with  an  angular  velocity  ooz  =  —  60  radians  per 
sec,  and  its  centre  has  a  horizontal  velocity  of  translation  of  1^=  50 
ft.  per  sec.  in  a  direction  making  an  angle  of  36°  52'  with  the  axis 
of  rotation.    Find  the  motion  of  the  ball. 

Ans.  Let  the  plane  of  ZX  be  horizontal  and  take  the  centre  as  origin.   Then, 

since  V  is  in  this  plane,  we  have 

Fx  =  +  30ft.  persec,      Vy  =  0, 
Fz  =  -|-  40  ft.  per  sec. 
Also, 
cox  =  0,     GJj/  =  0,     ooz=  —  60  rad.  per  sec. 

The  rotation  is  then  clockwise,  or  from 
Y  towards  X,  as  shown  in  the  figure. 

Then,  just  as  in  the  preceding  example, 
the  central  axis  is  parallel  to  the  axis  of  Z, 
and  the  position  of  the  central  axis  is,  from 
equation  (11),  page  203,  given  by 

x"  =  Q,    y'  =01=  -l^it.,     2"  =  0. 

If  then  we  neglect  the  acceleration  due  to  the  attraction  of  the  earth,  the 
motion  of  the  ball  is  a  screw  motion  consisting  of  a  velocity  uz  =  +  40  ft.  per 
sec.  along  the  axis  of  rotation  OZ  through  the  centre  0  of  the  biill,  and  a  rota- 
tion of  caz  =  —  60  radians  per  sec.  about  this  axis,  together  with  a  translation 
of  this  axis  of  Vx  =  +40  ft.  per  sec. 

Or,  neglecting  the  acceleration  due  to  gravity,  the  motion  is  the  same  (page 

193)  as  if  the  ball  were  part  of  a  cylinder  of  radius  01=  -^  ft.  whose  axis  OZ 

a 

is  the  axis  of  rotation  of  the  ball,  and  this  cylinder  rolls  on  the  horizontal  plane 
HIH  with  angular  velocity  gjz  =  —  60  radians  per  sec,  while  at  the  same  time 
the  cylinder  is  translated  along  OZ  with  the  velocity  -f-  40  ft.  per  sec 

The  centre  0  of  the  ball  moves  then  in  the  resultant  of  V-  and  Vz ,  or  along 
the  straight  line  OF  in  the  horizontal  plane  XZ,  with  a  velocity  F=  50  ft.  per 
sec,  at  an  angle  of  36°  52'  with  the  axis  of  rotation  OZ. 

If  now,  owing  to  gravity,  the  ball  falls  vertically  while  the  centre  moves 
along  OF,  then  we  must  consider  the  plane  HIH sls  falling  vertically  with  the 
ball.  The  centre  moves  then  in  a  curve  Oab,  the  projection  of  which  upon  the 
plane  XZ  is  a  straight  line  Oc. 

(21)  Ball-players  assert  that  the  projection  of  this  curve  Oab 
{preceding  example)  upon  the  plane  XZ  is  not  a  straight  line  but  a 
curve.     Explain  hoiv  this  can  be. 


COz 


Ans.  We  have  seen  in  the  preceding  example  that  if  the  centre  0  of  the 
ball  has  a  velocity  Fand  at  the  same  time  the  ball  has  an  angular  velocity  ooz 


CHAP.  II.]  RELATIVE   MOTION   OF   A   BODY. 


213 


around  an  axis  OZ,  tlie  centre  0  moves  with  the  velocity  Vz  along  the  axis  and 
at  the  same  time  the  axis  itself  moves  with  the  velocity  Vx,  Vz  and  Vx  beinff 
the  components  of  V  along  and  perpendicular  to  the  axis.  At  the  same  time 
the  ball  rotates  about  the  axis  OZ.  The  motion  of  0  is  then  in  the  straight 
line  OV.  ,  ° 

But  no  account  has  been  taken  of  the  resistance  of  the  air.  The  air  acts  to 
cause  a  retardation  of  Vz  and  Vx . 

If  the  retardation  in  each  case  were  proportional  to  the  velocity,  we  should 
still  have  motion  of  0  in  the  straight  line  0  V.  But  the  retardation  in  each 
case  is  not  proportional  to  the  velocities  but  more  nearly  proportional  to  the 
squares  of  the  velocities.  Hence  the  greater  component  is  retarded  propor- 
tionally more  than  the  less. 

If  then  the  rotation  axis  OZ  makes  an  angle  less  than  45°  with  the  direc- 
tion of  V,  Vz  is  greater  than  Vx  and  is  therefore  retarded  proportionally  more 
than  Vx.     The  centre  0  moves  then  in  an  "  out-curve  "  OB. 

If,  however,  the  axis  of  rotation  OZ 
makes  an  angle  greater  than  45°  with  the 
direction  of  F,  Fa;  is  greater  than  Vz  and 
is  therefore  retarded  proportionally  more 
than  Vz.  The  centre  0  moves  then  in  an 
"  incurve." 

In  either  case  the  velocity  is  retarded 
least  in  the  direction  of  least  resistance 
and  the  centre  swerves  in  the  direction  of 
the  smallest  component  of  V. 

Thus  by  "twisting  "  the  ball  the  pitcher  is  able  to  make  it  curve  slightly  by 
either  to  right  or  to  left  according  as  the  axis  of  rotation  makes  an  angle  with 
the  velocity  of  projection  greater  or  less  than  45°. 

If  the  axis  of  rotation  makes  an  angle  of  45°  with  the  velocity  of  projection, 
there  should  be  no  curve.  If  it  is  at  right  angles  to  the  velocity  of  projection, 
there  should  be  no  curve. 

The  cause  of  curvature  is  thus  due  to  the  resistance  of  the  air,  but  it  is  not, 
as  is  generally  supposed,  due  to  the  ball  rolling  upon  a  cushion  of  compressed 
air  in  front  of  it,  since  in  that  case  we  should  always  have  curvature  in  one 
direction  for  one  direction  of  rotation. 

In  the  first  of  our  figures  preceding,  such  action  tends  to  increase  the  "  out- 
curve."  But  in  the  second  it  tends  to  decrease  the  "  in-curve."  The  "  in-curve" 
would  not  be  possible  if  this  action  were  the  only  cause  of  curvature.  It 
ought  to  be  less  than  the  out-curve,  so  far  as  this  action  is  effective,  in  the 
figures  given. 

If  we  have  rotation  in  the  opposite  direction  from  that  in  the  figure,  or  if  the 
line  representative  of  ojz  is  positive  instead  of  negative,  the  rolling  of  the  ball, 
if  any,  upon  a  cushion  of  compressed  air  in  front  of  it  would  act  to  decrease 
the  ' '  out-curve  "  and  increase  the  ' '  in-curve. " 

Relative  Motion  of  a  Body. — When  a  body  at  any  instant  has 
two  simultaneous  motions  we  can  consider  the  body  itself  as  having 
one  of  these  motions  and  the  space  occupied  by  the  body  as  having 
the  other.  The  first  motion  is  then  that  which  the  body  would 
appear  to  have  to  an  observer  in  space  moving  with  space  and 
unaware  of  his  own  motion.  We  call  it  therefore  the  relative 
motion  of  the  body  with  reference  to  moving  space. 

We  have  thus  far  seen  how  to  determine  the  actual  motion  when 
we  have  given  the  relative  motion  and  the  motion  of  space.  We 
have  now  to  consider  the  inverse  problem  of  how  to  determine  the 
relative  motion  when  we  have  given  the  actual  motion  and  the  mo- 
tion of  space. 

We  can  solve  the  problem  in  two  ways.  We  can  resolve  the 
given  actual  motion  into  two  component  motions  one  of  which 
coincides  with  the  given  motion  of  space.  Then  the  other  must  be 
the  relative  motion  required.  Or  we  can  add  to  the  actual  motion, 
composed  of  these  two  component  motions,  a  third  motion  equal 


214 


EIGID  SYSTEM — TKANSLATION  AND  KOTATION.    [CHAP.  II. 


and  opposite  to  the  given  motion  of  space.     This  will  counteract 
one  of  the  components  and  leave  as  a  result  the  relative  motion. 

Relative  Motion  of  a  Body  with  Reference  to  Space  Transla- 
tion. — If  the  actual  motion  of  a  body  and  the  motion  of  the  space 
occupied  by  it  are  both  motions  of  translation,  the  relative  motion 
will  be  one  of  translation  also.  In  such  case  we  can  treat  the  body 
and  the  space  occupied  by  it  as  points,  and  thus  have  simulv  to 
find  the  relative  motion  of  a  particle  with  reference  to  the  point  of 
moving  space  occupied  by  the  particle. 

The  relative  velocity  of  the  particle  is  then  the  resultant  of  the 
actual  velocity  of  the  particle  and  the  velocity  of  the  point  of  mov- 
ing space  occupied  by  the  particle,  taken  as  acting  with  reversed 
direction. 

If  then  the  actual  velocity  of  the  particle  is  zero,  the  relative 
velocity  at  any  instant  is  always  equal  and  opposite  to  that  of"  the 
point  of  moving  space  occupied  by  the  particle  at  that  instant. 

Thus  if  the  particle  P  de- 
scribes an  ellipse  with  reference 
to  the  fixed  point  O  at  one 
focus,  the  relative  velocity  of 
O  with  reference  to  P  will  be 
always  equal  and  opposite  to 
the  velocity  of  P  at  any  in- 
stant, and  the  apparent  path  of 
O  as  seen  from  P  will  be  a 
similar  ellipse  with  P  at  a 
focus. 

Relative  Motion  of  a  Body 
with  Reference  to  Moving  Space 
in  General. — Any  motion  of  a 
body  at  any  instant  can  be  re- 
solved into  a  translation  of  any  point,  and  a  rotation  about  an  axis 
through  that  point  (page  190).  If  we  take  for  this  point  the  point 
of  space  occupied  by  any  particle  of  the  body,  we  have  translation 
only  of  this  point  and  particle,  and  the  relative  velocity  is  found 
as  in  the  preceding  Article. 

The  relative  velocity  of  the  particle  is  then,  as  before,  the  result- 
ant of  the  actual  velocity  of  the  particle  and  the  velocity  of  the 
point  of  moving  space  occupied  by  the  particle,  taken  as  acting 
with  reversed  direction. 

We  obtain  then  the  relative 
path  by  giving  to  the  actual  path 
the  reversed  motion  of  space. 

For  example,  let  the  actual 
velocity  of  a  particle  P  be  uni- 
form and  equal  to  c,  and  its  con- 
stant direction  be  in  the  direc- 
tion AB.  Let  the  line  AB  be 
the  diameter  of  a  circular  disk 
which  rotates  clockwise  with 
constant  angular  velocity  o 
about  the  axis  at  C.  We  obtain 
the  path  relative  to  the  disk  by 
supposing  rotation  of  the  actual 
path  AB  counter-clockwise. 

Thus  at  the  end  of  any  time 
t  the  particle  has  traversed  the  distance  AN  =  ct  and  the 


has  turned  clockwise  through  the  angle  oot. 


disk 
The  corresponding 


CHAP.  II.]  ACCELERATION   OF   RELATIVE   MOTION. 


215 


jelative  position  P  of  the  particle  is  then 
would  occupy  if  the  line  AB  were  turned 
<30unter-clock;wise  through  the  angle  cot. 
Repeating  the  construction  for  successive 
values  of  t  we  obtain  the  relative  path 
APCD.  The  end  D  corresponds  to  the 
rotation  angle  ooT,  where  T  is  the  time  of 
the  actual  motion  from  Ato  B.  It  goT  =  Tt, 
the  point  D  would  coincide  with  A.  Let  r 
be  the  radius  of  the  disk.  Then  from  the 
two  equations  cT  =  2r,  and  ooT=it,  we 
have  for  the  condition  of  this  coincidence 
of  D  and  A, 

oa        It 

If  the  actual  path  AB  makes  an  acute 

angle  with  the  axis  of  rotation  through  C, 

the  relative  path  lies  on  the  surface  of  a  cone. 

Acceleration  of  Relative   Motion. — Let  a  particle  describe  the 

path  MN  with  any  motion,  and  at  the  same  time  let  this  path  have 

a  motion  of  translation.  Then  we  can 
regard  the  first  motion  as  relative 
with  reference  to  the  second,  and  its 
acceleration  /i  is  the  relative  accelera- 
tion. Besides  this  relative  acceleration 
at  any  instant,  the  particle  has  the 
acceleration  /a  of  the  motion  of  trans- 
lation at  that  instant.  The  actual 
acceleration  of  the  particle  is  then  the 
resultant  of  the  two  accelerations  fi 
and  fi. 
It  is,  however,  different  when  the  path  MN  has  any  motion  in 

general,  because  such  motion  may  be  resolved  into  a  motion  of 

translation  of  any  point  of  the  path  and  a  motion  of  rotation 

about  an  axis  through  that  point. 
Take  for  this  point  the  point  of 

space  occupied  by  the  particle  at  any 

instant. 

Then  we  have  besides  the  accelera- 
tion /i  of  the  particle  in  its  path,  and 

the  acceleration  /a  of  the  point  of 

space   occupied  by   the   particle,   a 

third    acceleration,  /s,   due    to    the 

rotation,  which  we  can  determine  as 

follows:  ^    ,,  X-    ^  rni-  • 

Let  V  be  the  relative  velocity  of  the  particle.  Then  in  an  in- 
definitely small  time  dt,  vdt  will  be  the  element  MN  of  the  relative 
path  This  element  in  the  time  dt  is  translated  to  PQ  and  at  the 
same  time  has  the  angular  velocity  oo  about  the  point  of  space  oc- 

^"^Let  the  axis  OP  through  this  point  piake  the  angle  6  with  P^. 
Then  at  the  end  of  the  time  dt,  4>  will  be  at  R.  If  .A  is  the  accelera^ 
tion  in  the  direction  QR,  then  the  distance  QR  will  be 


QB  =  ^f»dt\ 


216 


EIGID  SYSTEM — TRANSLATION  AND  ROTATION.    [CHAP.  II. 


The  radius  of  rotation  is  OQ  =  vdt  sin  fi.  Hence  the  distance  QR 
is  also  given  by 

QR  =  OQ .  codt  =  vdt  sin  0 .  oadt. 

Equating  these  two  values  of  QR,  we  obtain 

/$  =  2vw  sin  Q. 

Hence  we  see  that  the  actual  acceleration  of  the  particle  in 
general  is  the  resultant  of  three  accelerations : 

The  first, /i,  is  the  acceleration  of  the  relative  motion  of  the 
particle. 

The  second,  fa  is  the  acceleration  of  the  point  of  space  occupied 
by  the  particle. 

The  third,  fa,  is  equal  in  magnitude  to  twice  the  product  of  the 
relative  velocity  v  of  the  particle,  the  angular  velocity  oo  of  the 
point  of  space  occupied  by  the  particle,  and  the  sine  of  the  angle  0 
which  the  element  of  the  relative  path  makes  with  the  axis  through 
the  point  of  space  occupied  by  the  particle.  Its  direction  is  at 
right  angles  to  the  plane  of  this  axis  and  element,  and  it  acts  in  the 
direction  given  by  the  rotation. 

If  then  fi,ft,  fa.  Fig.  1,  represent  these  accelerations,  we  have 
by  completmg  the  polygon  in  Fig.  2  the  actual  acceleration  /. 


Fig.  1. 


Fig.  3. 


Inversely,  if/,  fa  and  fa  are  given  and  it  is  required  to  find  the 

relative  acceleration /i,  we  must  take /a  and /a  reversed  in  direction. 

Illustrations. — Let  a  particle  P  move  at  any  instant  with  the 

velocity  v  in  the  direction  of  a  diameter  of  a  circular  disk.    Let  the 

disk  at  this  instant  have  an  angular 

velocity  oo  about  its  axis  at  C,  and 

the  distance  CP  of  the  particle  from 

the  axis  be  x. 

Then  the  relative  acceleration  is 

dv 
fi  —  -^j  along  the  diameter.    The 


dt 
acceleration  of  the  point  of  space 
occupied  by  the  particle  is /a  =  xoo"", 
along  the  diameter.  This  is  the 
central  acceleration  of  the  point  P 
due  to  rotation  about  C.  We  have 
also  /a  =  2voo,  acting  at  right  angles 
to  the  plane  of  the  element  of  the 
relative  path  and  the  axis  through 
P  parallel  to  the  axis  at  C,  and  it  acts  in  the  direction  given  by  the 
rotation  as  shown.  The  angle  9  which  the  element  of  the  path 
makes  with  the  parallel  axis  at  P  is  90°,  and  hence  sin  0  =  1.  The 
actual  acceleration  is  the  resultant  of  these  three  accelerations. 


CHAP.  II.] 


ILLUSTKATIONS. 


217 


Let  a  particle  P  move  at  any  instant  with  the  velocity  v  in  the 
circumference  of  a  circle  of  radius  r.  about  the  centre  d,  and  at 
the  same  time  let  the  centre  Ci  revolve  about  the  point  C»  with  the 
angular  velocity  go.    Let  the  distance  of  P  from  Ca  be  r^. 


Then  the  relative  acceleration /i  is  the  resultant  of  the  tangential 
dv  v^ 

acceleration  -:—  and  the  central  acceleration  —  acting  towards  Ci. 
at  ri 

The  acceleration  /a  of  the  point  of  space  occupied  by  the  particle 
is  n&j'  acting  towards  Ca.  We  have  also  /s  =  2va)  acting  at  right 
angles  to  the  plane  of  the  element  of  the  relative  path  and  the  axis 
through  P  parallel  to  the  axis  at  Ca,  and  it  acts  in  the  direction 


J^=rco8X.co 


^7i=2»w«fnA 


given  by  the  rotation  as  shown.    The  actual  acceleration  is  the 
resultant  of  these  three  accelerations. 


818  EIGID  SYSTEM — TRANSLATION  AND  EOTATION.    [CHAP.  II. 

Let  a  particle  P  move  on  a  meridian  of  the  earth  and  have  at 
any  instant  the  velocity  v  and  the  tangential  acceleration     ,  .     If  r 

is  the  radius  of  the  earth,  the  central  acceleration  is  —  and  the 

r 

relative  acceleration /i  is  the  resultant  of  -rr  and  — . 

If  A  is  the  latitude  of  P,  r  cos  A.  is  the  radius  of  rotation ;  and  if  &> 
is  the  angular  velocity  of  the  earth,  the  acceleration  ^  of  the  point 
of  space  occupied  by  the  particle  is/2  =  rcosA  .  &5^ 

We  have  also  fa  —  2voo  sin  A  acting  at  right  angles  to  the  plane  of 
the  element  of  the  relative  path  and  the  axis  through  P  parallel  to 
the  earth's  axis,  that  is,  tangent  to  the  latitude  circle  at  P.  It  acts 
towards  the  east.  The  actual  acceleration  is  the  resultant  of  these 
three  accelerations. 


CHAPTEK  in. 


GENERAL  ANALYTICAL  RELATIONS  FOR  A  POINT  OF  A 
RIGID  ROTATING  SYSTEM.  EULER'S  GEOMETRIC 
EQUATIONS. 

General  Analytical  Relations  for  a  Point  of  a  Rigid  Rotating 
System. — Let  a  rigid  system  rotate  at  any  instant  about  the  axis 
IC  with  the  angular  velocity  co  and  the  angular  acceleration  a. 
Take  any  point  O  of  the  system  as  origin,  and  let  the  direction- 
cosines  of  fij  be  cos  a,  cos  p,  cos  y. 


Then  we  have  for  the  components  of  co  and  a 
cox=  oa  COS  a,  a)y=  00  COS  /?,  coz  =  co  COS  y ; 
ax=  oc  COS  a,      ay=  a  COS  /?,     ««  =  a  COS  y ; 

COS  a  —  —  =  — ,     COS  P  = =  — -,     COS  y  —        — 

00  a  00         a  <» 

and  since  cos''  a  +  cos''  /3  +  cos"  y  =  1, 


(1) 


(2) 


Let  (x  «,  2)  be  the  co-ordinates  of  any  pomt  on  the  axis  IC,  and 
<x'  V  z')  the  co-ordinates  of  any  point  P  whose  distance  from  the 
axis  is  PC  =  r.    Then,  as  we  have  seen  (page  190),  we  can  resolve 

219 


220  POINT  OF  RIGID  ROTATING  SYSTEM  :  GENERAL  RELATIONS.  [CH.III. 


the  rotation  about  IC  into  an  equal  angular  velocity  about  a  paral- 
lel axis  through  the  origin  O  and  a  velocity  u  =  tap  of  the  origin  O, 
where  p  is  the  distance  10  of  the  origin  from  the  axis. 

The  components  of  this  velocity  of  O  are  as  on  page  179,  equa- 
tion (1): 


Vx  =  oozy  —  00  yZ ; 

Vy  =  OOxZ  —  COzX ; 

Vz  =  oOyX  —  ajxy. 


(3) 


The  components  of  the  linear  velocity  v'  of  P  due  to  rotation 
about  the  parallel  axis  through  the  origin  O  are  as  on  page  198^ 
equations  (6) : 

v'x  =  ooyz  —  cozy' ;  1 

V'y  =  cozx'  —  coxZ' ;   I (4) 

V'z  —  ODxy'  —  OOyX.    J 

We  have  thus  the  total  components  of  the  velocity  P,  just  as  on. 
page  198,  equation  (7): 

Vx=vx+  v'x ;  1 

Vy=Vy+   V'y\     \ (6> 

Vz  =  Vz-\-  v'z. 

The  components  of  the  linear  tangential  acceleration  of  P  due 
to  rotation  about  the  parallel  axis  through  O,  we  see  from  (4),  are 
given  by 

ftx  =  ocyZ'  —  azy  ; 


fty  =  azX'  —  axZ  ;    ■ 
f  tz  =  ocxyf  —  ayX  . 
Since  poo  =  v  and  'p'^oo^  =  v'^  —  Vx^  +  %"  +  r/,  we  have  from  (3) 

J  _  (oozy  —  coyZf       (coxZ  —  cazXf        (ooyX  —  ooxVf 


(6) 


(7) 


Let  f'n  be  the  normal  linear  acceleration  of  the  point  P  due  to 
rotation  about  the  parallel  axis  through  the  origin  O.  Then  f'n  = 
v'oo ;  and  since  velocity  in  the  hodograph  is  the  normal  acceleration 
in  the  path  (page  52),  we  have  directly  from  the  figure,  for  the  com- 
ponents otj'n, 

fvx  =  v'zWi/  —  V',,a)z  ;    1 

fny  =  t^'xooi  —  v'zGOx  :,      > (^) 

J  nz  =^  V  yOOx  —  V  xOO^.      J 

We  have  then  for  the  components  of  the  acceleration  /'  of  the 
point  P,  from  (8)  and  (6), 

fx  =fnx  +ftx  =  {V'zooy  —  V'ycoz)  +  (ayZ'  —  azy')]    '] 

fy  =fnv  +fty  =  (v'xooz  —  v'zOOx)   +  {cczX  —  axZ')\    \    -      •     (9) 

fz  =f'nz  +  f'tZ  =  {V'yax  -  V'xCOy)    +  {fiCxlf  -  UyZ').     J 


CHAP.  III.]  KINEMATICS   OF   A   RIGID   SYSTEM.  221 

If  we  put  for  v'x,  v'y,  v'z  their  values  as  given  in  (4),  we  have 

o'  —  axZ')  ;    I 
fz  =  i.ooxX'  +  ooyy'  +  a)zZ)oo^  —  aPz    +  {axV   —  ocyZf).    J 


fx  =  (cJjrCC'  +  oOyy'  +  (iazZ')oox  —  oorx'  +  {cxyz'  —  azy); 

f'y  =  {ojxJC  +  GOyy'  +  OL)zZ')coy  —  oaY  +  {oczX'  —  axZ') ;     I  ■      •     (10) 


Equations  (10)  give  the  valueg  of  the  components  of  the  linear 
acceleration/  of  any  point  P  of  the  system  whose  co-ordinates  are 
(x',  y',  z'),  in  terms  of  these  co-ordinates  and  the  components  of  a> 
and  a. 

The  moments  of  the  component  linear  accelerations  with  refer- 
ence to  the  origin  O  are 


fxVy'  +  2f\   fy  Vx"  +  z!\   fz\^x'^  +  y\  .    .    .    (11) 

For  the  moments  about  the  axes  of  the  components  of  rf  and  /' 
we  have: 

about  X  parallel  to  plane  YZ,  Mx  =  tfzy'—  Vyz',  or  fzy  -fyZ' ;  1 
"      Y       "        "      "     ZX,  My  =  if xz'—v'zx,  or  fxZ—f;^;Ui3) 
"      Z      "        "      "     XY,Mz^v'yXf-v'xy',  or  fyx-fxy'.] 
The  resultant  moment  in  both  cases  is  given  by 


Mr=VMx''  +  My^  +  Mz'' (13) 

Its  line  representative  has  the  direction-cosines 

(14) 


Mx  My         Mz 


My'       Mr'      Mr 

Looking  along  the  line  representative  towards  the  origin  the  rota- 
tion is  counter-clockwise. 

[We  can  deduce  equations  (9)  directly  by  the  Calculus.     Thus  if  we  dif- 
ferentiate the  values  of  v'x,  i>'y,  i>  z  given  by  (4),  then,  since 


and 


we  have  at  once 


dt='""  ■df='^-  Tt-"^'' 

dwx  dooy  dooz 

——-  =  ax,  --JT-  =  ocy,  —^  =  az, 
dt                   dt  dt 


f^  =  -^  =  {v'zcoy  -  vycoz)  -f  {ayz'  -  azjH; 
at 

fy  =  -^  =  (i/xa>z  -  v'zoox)  -f  {azx'  -  aa«'); 

fg  =  ^  =  {vycox  -  v'xGOy)  -f  {axy'  -  ayz'). 

Euler's  Geometrical  Equations.— To  determine  the  geometrical 
equations  beticeen  the  motion  of  a  rigid  system  m  space  and  th^ 
angular  velocity  of  the  system  about  an  axis  in  the  system. 


222  EULER's   geometrical   equations.  [chap.  III. 

Let  OXi,  OYi,  OZi  be  rectangular  co-ordinate  axes,  fixed  in  the 
system  and  therefore  rotating  with  it,  and  let  the  system  rotate 
about  some  axis  fixed  in  the  system,  and  therefore  making  in- 
variable angles  with  these  axes, 
so  that  the  component  angular 
velocities  in  the  co-ordinate 
planes  are  (»a:, ,  o^j/, ,  «»2,.  We 
take  direction  of  rotation  as  al- 
ways 

about  X  from  F.  to  Z,  1  1*^'*'!^' 
the     op- 

"     F.     "     Z,  "X  !>positedi- 

"     Zi     "      Jf  "  F    '  rection 
'        '  J  negative. 

Let  now  OX,  OY,  OZhe  rect- 
angular co-ordinate  axes  whose 
directions  m  space  are  invariable. 
For  instance,  the  axis  OZ  may 
be  always  directed  towards  the 
North  Pole,  then  XY  is  the  plane  of  the  celestial  equator. 

Let  the  point  O  be  taken  as  the  centre  of  a  sphere  of  radius  r. 
Let  X,  F,  Z  and  Xi ,  Fi ,  Zi  be  the  points  in  which  this  sphere  is 
pierced  by  the  fixed  and  moving  axes. 

Let  the  axes  OXi ,  OF  ,  OZi  have  the  initial  positions  OX,  OY, 
OZ.  First  turn  the  system  about  OZ  as  an  axis  through  the  angle 
XZP  =  rp,  so  that  OX  moves  to  OP,  and  OF  to  OD.  Then  turn  the 
system  about  OD  as  an  axis  through  the  angle  ZOZi  —  6,  so  that  OP 
moves  to  OE,  and  OZ  to  OZu  Finally  turn  the  system  about  OZi 
as  an  axis  through  the  angle  EZXi  =  0,  so  that  OE  moves  to  OXi  , 
and  OD  to  OF. 

It  is  required  to  find  the  geometric  relations  between  6,  cp,  ^  and 
GOxj ,  00 y.,  coz^  as  the  system  rotates.  These  geometric  relations  are 
called  Euler's  Geometric  Equations. 

Let  the  angular  velocity  of  Zi  perpendicular  to  the  plane  ZOZi 

at  any  instant  be  denoted  by  ~-.   This  is  called  the  angular  velocity 

of  precession.    Let  the  angular  velocity  of  Zi  along  ZZi  at  the  same 

do 
instant  be  denoted  by   . , .    This  is  called  the  angular  velocity  of  nu- 
tation.    Let  the  angular  velocity  of  Xi  with  reference  to  E  at  that 

instant  be  denoted  by  ^v. 
at 

Draw  ZiN  perpendicular  to  OZ.     Then  ZiN  =  r  sin  0,  and  the 

linear  velocity  at  any  instant  of  Z,  perpendicular  to  the  plane  ZOZ\ 

M  do 

is  r  sin  0  .  -^,  and  along  ZZi  at  the  same  instant  it  is  r     .  The  linear 

dt  at 

velocity  at  the  same  instant  of  Zi  along  FiZi  is  ra>xi ,  and  along  ZiXi 
it  is  ro),,^. 

We  have  then  directly  from  the  figure 

d^  ^  ,  •    ^ 

r^^  =  rooy^  cos  0  -f-  7-00x1  sin  0 ; 

r  sm  e  .  -— -  =  rooy.  sm  0  —  rco^,  cos  <p. 
at 


CHAP.  III.] 


KINEMATICS  OF  A  RIGID  SYSTEM. 


Since  the  radius  r  cancels  out, 
do 


jT  =  ooyi  COS  (p  +  ooxj  sin  (f>; 

dtl^ 
sm  9  ^  =  (»j,,  sin  0  —  GJx,  COS  (p. 


223^ 


(1> 


Combining  these  two  equations, 


de 


d^ 


<»a;,  =  ^  sin  0  —  ^  sin  e  cos  0  ; 
ooy^  =  ^  COS  0  +  -5^  sin  S  sm  0. 


(2> 


In  the  same  way  by  drawing  a  perpendicular  from  E  to  OZOE 
we  have  the  linear  velocity  of  E  perpendicular  to  ZOE  equal  to 

r  cos  e  ^,  and  of  Xi  relative  to  E  along  EX^ ,  r^. 
etc  d^ 

The  whole  velocity  of  X,  in  space  along  X,  Yi  is  rcog,.    Hence 


dip        .      d0 

'"^i  =  TT*  cose  +  -5- 

dt  dt 

Equations  (1),  (2)  and  (3)  are  Euler's  Greometric  Equations. 


(3> 


EXAMPLES. 


(1)  Deduce  the  angular  velocities  oox,  coy,  ooz  about  the  fixed  axiSy 
in  terms  of  6,  0,  rp. 

Ans.  Let  co,-  be  tlie  resultant  angular  velocity  about  the  fixed  axes.  If 
we  impress  on  space  and  also  on  the  system,  in  addition  to  its  existing 
motion,  an  angular  velocity  equal  to  —  a?,-  about  the  resultant  axis  of  rotation, 
the  axes  OXi,  OYi,  OZi  will  become  fixed  and  OX,  OY,  OZ  will  move  with 
angular  velocities  —  oox ,  —  coy ,  —  coz.  Hence  in  the  equations  already 
found  we  have  only  to  replace  0  by  —  V.  S  by  —  6,  ^  by  —  0,  and  coxi ,  <»j/, , 
cozj  will  become  —  cox ,  —  ooy ,  —  coz ,  and  we  have 


dO  .     ,    .  d(p    .    .         , 

oox=  —  T-  sm  tp  +  -J-  sm  6  cos  ip; 
at  dt 

dB  ,    ,  ^0    •    o    •     I 
GOu  =  ^-  COS  il>  4-  —rr  sm  0  sm  ^ ; 
dt  dt 

d(p  o   ■    <^^ 


(4) 


(2)  Refer  the  axes  fixed  in  space  to  the  axes  fixed  in  the  system. 

Ans  "We  have  simplv  to  Interchange  in  the  figure  Xi,  Yi,  Zx  with  X,  F, 
Z  each  with  each.  If  then  the  angles  Q,  0,  ^  are  still  measured  as  indicated 
in  the  figure,  the  relations  connecting  them  with  the  angular  velocities  are  ob- 
tained by  changing  cox,,  coy,,  coz,  into  —  a)x,  -  o^y  <  -  Mf-     .    ^,     „  ., 

If  we  measure  Q  in  the  direction  opposite  to  that  mdicated  in  the  fagure,  the 
expressions  for  oox,  ony  are  identical  with  those  already  found  for  <»ar„  ooy,. 


224 


EULER'S   geometrical   equations.  [chap.  III. 


(3)  If  p,  q,  r,  are  the  direction-cosines  of  OZ  with  regard  to  the 
axes  OXi ,  OFi ,  OZi ,  show  that 


-^^  -  Qoozr  +  rooy,  =  0; 


dq 
di 


—  rooxt  +  poozi 


dr  , 

--  -  pooy,  +  qoox. 


(5) 


Ans.  Any  one  of  these  may  be  obtained  by  differentiating  one  of  the  ex- 
pressions 

p  =  —  sin  6  cos  0,     g'  =  sin  0  sin  (p,     r  ■=  cos  9 

(4)  Show  that  the  direction-cosines  of  either  set  of  Euler^s  axes 
with  regard  to  the  other  are  given  by 

cos  XXi  =  —  sin  tJ;  sin  (p  -\-  cos  ip  cos  (p  cos  6;    ] 

cos  YXi  =  cos  ^  sin  0  -|-  sin  tf>  cos  <p  cos  6;         V     ...     (6) 

cos  ZXi  =  —  sin  0  cos  <p.  J 

cos  XYi  =  —  sin  rp  cos  0  —  cos  ^  sin  0  cos  6;  \ 

cos  YYi  =  cos  jp  cos  0  —  sin  ip  sin  0  cos  6;  >•     .     .     .     (7) 

-  cos  ZFi   =  sin  9  sin  0.  j 

cos  XZi   =  sin  S  cos  ip;  j 

cos  FZi   =  sin  9  sin  ^;   V   ...     .  (8) 

cos  ZZi    =  cos  0.  j 

Ans,  We  have  from  the  figure  the  following  spherical  triangles  for  which 
we  know  two  sides  and  the  included  angle  : 


Triangle.       Sides.  Angle. 

_  nX   =90°-4-j/» 
JDX:X  Dx,  =90-0    XDX,=B 


Triangle.    Sides.  Angle. 

D^y^  DX  =90  +  ^    Fi2>X=180-9 


Z,ZX,  ZZ^  =  9 


Z)FF,   Y'D=t  I^i2>r=180-e 

Z2'iX,=18O-0Z,ZFi  Ji2,  =90  ri^i2=9O-0 

ZiZ  =0 


Triangle. 


Angle. 
ZiPXi  =  90° 


PFZ, 


FPZ.    =  90 


Sides. 
PX  =il) 
PZ,  =90-0 

PZ,  =90-0 
PY  =  90  -  ^ 

Solving  these  triangles  we  have  at  once  equations  (6),  (7),  (8). 

(5)  Prove  in  the  same  way  the  following : 

cos  XiX  =  —  sin  ^  sin  0  +  cos  Tp  cos  0  cos  0;  \ 

cos  FiX=  —  sin  ip  cos  0  —  cos  ip  sin  0  cos  0;  > 

cos  ZiX  =  sin  9  cos  ^.  J 

cos  Xi  F  =  cos  ^  sin  0  +  sin  ^  cos  0  cos  9  ;  1 

cos  Fi  F  =  cos  ip  cos  0  —  sin  ^  sin  0  cos  9;  > 

cos  Zi  F  =  sin  0  sin  ^.  ) 


(9) 


(10) 


CHAP.  III.]  KINEMATICS   OF  A   RIGID   SYSTEM — ROTATION. 


225 


COS  XiZ  =  —  sin  6  cos  <p;  ^ 

cos  YiZ  =  sin  6  sin  0;         >■ (11) 

cos  ZiZ    =  cos  6.  \ 

(6)  Mnd  the  relations  between  the  co-ordinates  x,  y,  z,  of  the 
fixed  system  of  axes  and  the  co-ordinates  Xi,  yi,  Zi  of  the  moving 
system. 

Ans.  If  we  multiply  the  first  of  equations  (6)  by  x,  the  second  by  y,  the  third 
by  s  and  add,  and  do  the  same  for  (7)  and  (8),  we  have  at  once,  as  we  see  from 
the  figure, 

Xi  =  (—  sin  ^  sin  0  +  cos  rj}  cos  (p  cos  B)x 

-}-  (cos  rj)  sin  0  +  sin  ^  cos  (f>  cos  B)y  —  sin  6  cos  (p.z; 

yi  =  (—  sin  il>  cos  0  —  cos  ^  sin  (p  cos  S)z  }•   .     .     (12) 

-|-  (cos  i()  cos  0  —  sin  V'  sin  0  cos  Q)y  +  sin  9  sin  0.8; 

2i  =  sin  9  cos  iff.x  +  sin  9  sin  ^.y  -|-cos  9  .  s. 

In  the  same  way  we  have  from  equations  (9),  (10),  (11), 

X  =  ( —  sin  ip  sin  <p  +  cos  tfj  cos  <p  cos  Q)Xi 

-\-  (—  sin  if}  cos  <p  —  cos  V'  sin  0  cos  9)^i  4*  sin  9  cos  ^.Zi; 

y  =  (cos  ^  sin  0  4-  sin  ip  cos  0  cos  9)xi  }■  .    (13) 

-|-  (cos  ip  cos  0  —  sin  ip  sin  0  cos  6)yi  -)-  sin  9  sin  ^.0i; 

2  =  —  sin  6  cos  (p.Xi  -f-  sin  9  sin  (p.yi  -j-  cos  Q.Zu 


END  OF  VOLUME  I. 


MECHANICS. 


INDEX. 


Vol.  I. 
KINEMATICS. 

Absolute— motion,  13;  position,  10;  rest,  13. 

Acceleration — angular,  175;  composition  and  resolution  of,  176;  concurring 
angular  accelerations,  178;  couple,  181;  instantaneous  axis  of,  175;  line 
representative  of,  175;  moment  of,  177;  uniform  and  variable,  175;  unit 
of,  176;  and  translation,  composition  and  resolution  of,  189;  in  terms  of 
linear,  75,  176;  in  terms  of  moment  of  linear,  76. 

Acceleration — linear,  48,  92;  of  gravity,  9;  of  relative  motion,  215;  in  terms  of 
angular  acceleration,  75,  176;  central,  86,  92,  99,  103;  centre  of,  92;  line 
representative  of,  49;  mean  and  instantaneous,  48;  moment  of,  60;  moment 
of,  in  terms  of  angular,  76;  normal,  in  terms  of  angular,  76;  proportional 
to  force,  91;  paracentric,  87;  resolution  and  composition  of,  49;  resultant 
of,  50,  63;  sign  of  components  of,  50;  tangential  and  normal,  52;  triangle 
and  polygon  of,  49;  unit  of,  49;  uniform  and  variable,  49;  uniform,  in- 
clined to  direction  of  motion,  117. 

Amplitude  of  an  oscillation,  104. 

Analytical  relations  for  a  rigid  rotating  system,  319. 

Angle,  conical,  unit  of,  6;  solid,  7;  unit  of,  5. 

Angular — displacement,  171;  composition  and  resolution  of,  171;  concurring 
angular  displacements,  178;  couple,  181;  line  representative  of ,  170;  rigid 
system,  170;  in  terms  of  linear,  170;  and  translation,  composition  and  reso- 
lution of,  187. 

Angular  revolution  of  a  point,  71. 

Angular — speed,  72;  mean  and  instantaneous,  71;  numeric  equations  of,  72; 
sign  of,  72;  unit  of,  72;  in  terms  of  linear  velocity,  74;  iu  terms  of  moment 
of  velocity,  75;  in  terms  of  normal  acceleration,  76;  rate  of  change  of,  73; 
numeric  equations  of  rate  of  change  of,  73;  sign  of  rate  of  change  of,  73; 
equations  of  motion  under  different  rates  of  change  of,  73;  rate  of  change 
of,  in  terras  of  linear  speed,  75;  rate  of  change  of,  in  terms  of  moment 
of  tangential  acceleration,  76;  graphic  representation  of  rate  of  change 
of,  77. 

Angular — velocity,  174;  mean  and  instantaneous,  174;  unit  of,  174;  uniform 
and  variable,  175;  of  a  rigid  system,  174;  composition  and  resolution  of, 
176;  in  terms  of  linear  velocity,  176;  moment  of,  177;  concurring  angular 
velocities,  178;  resultant  of,  178;  couple,  181;  and  translation,  composition 
and  resolution  of,  189;  centre  of  parallel  angular  velocities,  192;  result- 
ant, for  a  rigid  system,  197. 

227 


228  INDEX. 

Area,  measures  of,  9. 

Axis,  of  earth,  length  of,  5,  9;  of  rotation,  173;  of  instantaneous  rotation,  174; 
instantaneous,  of  angular  acceleration,  175;  central,  190;  intersecting,  ro- 
tation about,  195. 

Blackburn's  pendulum,  137. 

Body,  material,  1;  projected  up  or  down,  93;  rigid,  170;  relative  motion  of,  218. 

Brachistochrone,  158. 

Cartesian  co-ordinates,  11. 

Centimeter,  value  of,  9. 

Central,  acceleration,  86,  92,  99,  103;  axis,  190. 

Centre,  of  acceleration,  92;  of  moments,  60;  of  angular  velocities,  192. 

Circle,  curvature  of,  7;  motion  in,  76,  154,  159. 

Coefficient  of  resistance  in  resisting  medium,  111,  113. 

Combined  rotation  and  translation,  13. 

Components,  of  displacement,  36;  of  motion  of  a  rigid  system,  201;  of  veloc- 
ity, 44. 

Composition  and  resolution,  of  accelerations,  49;  of  displacements,  36;  of  an- 
gular displacements,  171;  of  angular  velocities  and  accelerations,  176;  of 
translation  and  angular  displacements,  187;  of  translation  and  angular 
velocities  or  accelerations,  189;  of  moments, -62;  of  screws,  203;  of  veloci- 
ties, 43. 

Compound  harmonic  motion,  131;  resolution  and  composition  of,  131;  graphic 
representation  of,  135;  application  of  calculus,  137. 

Concurring  velocities,  accelerations  and  displacements,  resultant  of,  37,  45,  50, 
63-67;  angular  displacements,  velocities  and  accelerations,  178. 

Configuration,  12. 

Conical  angle,  6. 

Constrained  motion  of  a  point,  88,  151. 

Co-ordinates,  Cartesian,  10. 

Cosines,  direction,  12. 

Couple,  angular  displacement,  velocity,  acceleration,  181;  moment  of,  186. 

Curvature,  7;  unit  of,  8;  of  a  circle,  7. 

Curved  path,  motion  in,  153. 

Curvilinear  translation.  130. 

Cycloid,  motion  in,  155;  application  of  calculus,  157. 

Definition,  of  kinematics,  1,  15;  of  statics,  1;  of  mechanics,  1. 

Degree,  5. 

Derived  unit,  2;  dimensions  of,  3. 

Differential  equations  of  motions  of  a  point,  81. 

Dimensions,  of  a  derived  unit,  3;  of  space,  12;  of  unit  of  speed,  15;  of  unit  of 
rate  of  change  of  speed,  24. 

Direction  cosines,  12. 

Displacement,  .34;  line  representative  of,  34;  relative,  34,  35;  triangle  and  poly- 
gon of,  35;  composition  and  resolution  of,  36;  rectangular  components  of, 
36;  sign  of  components  of,  36;  resultant,  37;  moment  of,  60;  angular,  of  a 
rigid  system,  170;  line  representative  of,  170;  linear  in  terms  of  angular, 
170;  angular,  composition  and  resolution  of,  171;  angular,  concurring,  178; 
resultant  of,  178;  couple,  181;  of  a  rigid  system,  187;  composition  and 
resolution  of  translation  and  angular  displacement,  187. 

Dyne,  value  of,  9. 

Earth's  polar  axis,  length  of,  5. 

Epoch,  in  harmonic  motion,  105. 

Equations,  homogeneous,  3,  17;  of  speed,  16;  of  rate  of  change  of  speed,  25; 
of  motion  under  different  rates  of  change  of  speed,  27,  71;  of  motion  of  a 
point  under  different  accelerations,  50;  numeric,  of  angular  speed,  72;  nu- 
meric, of  rate  of  change  of  angular  speed,  73;  of  motion  under  different 
rates  of  change  of  angular  speed,  73;  differential,  of  motion  of  a  point,  81; 
for  falling  body,  93;  of  rotating  rigid  system,  176;  geometrical,  of  Euler, 
221 

Euler,  geometrical  equations  of,  321. 


VOL.   I. — KINEMATICS.  229 

Force  proportional  to  acceleration  91 
Falling  body,  93. 

<Jeometrical  equations  of  Euler,  221. 

Gram,  value  of,  9. 

Graphic  representation  of  rate  of  change  of  soeed  29 

Gravitation,  law  of,  99.  . 

Gravity,  acceleration  of,  9,  92. 

Harmonic  motion,  simple    103;    amplitude,  epoch,  period,  phase,  105-    com- 
^:%^s^TSZ;^^^-'  «^-  '''■'  ^-P^c^epre^iStioH, 
Hodograph,  53. 
Homogeneous  equations,  3,  17. 

Inclined  plane,  motion  on,  151. 

Instantaneous  acceleration,  48;  axis  of  rotation,  174;  axis  of  angular  accelera- 
locU     42'  ^       ^^  ""^  ^^^^'  ^'  ^^^^^'  ^^'  angular  speed,  72;  ve- 

Integral  curvature,  7. 
Intersecting  axes,  rotation  about,  195. 
Invariant,  203. 
Isochronous  oscillation,  104. 

Kepler's  laws,  140. 

Kilogram,  value  of,  9. 

Kinematics,  definition  of,  1,  15;  of  a  point,  91;  of  a  rigid  system,  169. 

Law  of  gravitation,  99. 

Laws  of  Kepler,  140. 

Length,  of  meridian,  5;  measures  of,  8;  of  polar  axis  of  earth,  5;  unit  of  4  8- 
displacement,  34;  velocity,  42;  acceleration,  48.  >    »    > 

Linear,  displacement  in  terms  of  angular,  170;  acceleration  in  terms  of  angu- 
lar, 176;  velocity  in  terms  of  angular,  176. 

Line  representative,  of  displacement,  34;  of  velocity,  43;  of  acceleration,  49;  of 
angular  displacement,  170;  of  angular  velocity,  174;  of  angular  accelera- 
tion, 175;  of  moment  of  displacement,  velocity  or  acceleration,  61,-  177. 

Mass,  measure  of,  4,  9;  unit  of,  4,  5. 

Material,  body,  1;  particle,  point,  14;  system,  14. 

Matter,  states  of,  1. 

Mean,  curvature,  7;  speed,  15;  rate  of  change  of  speed,  24r- Velocity,  42;  accel- 
eration, 48;  angular  speed,  72;  rate  of  change  of  angular  speed,  73;  angu- 
lar velocity,  174;  angular  acceleration,  175. 

Measurement,  2;  unit  of,  2. 

Measures,  table  of,  8;  of  length,  8;  of  area,  volume,  mass,  9. 

Mechanics,  definition  of,  1. 

Medium,  resisting — motion  in,  111;  coefiicient  of  resistance  in.  111,  113;  mo- 
tion of  projectiles  in,  127. 

Meridian,  length  of,  5;  relation  of,  to  meter,  5. 

Meter,  relation  of,  to  meridian,  5;  value  of,  9. 

Moment,  of  displacement,  velocity  or  acceleration,  60;  line  representative  of, 
61 ;  composition  and  resolution  of,  62;  sign  of,  62;  centre  of,  60;  linear  ve- 
locity in  terms  of  angul  \r  speed,  75;  of  tangential  acceleration  in  terms  of 
rate  of  change  of  angular  speed,  76;  of  angular  velocity  or  acceleration, 
177;  of  a  couple,  186. 

Motion,  13;  of  translation,  13;  of  rotation,  13;  equations  of ,  under  different  rates 
of  change  of  speed,  27;  equations  of,  under  different  accelerations,  50;  equa- 
tions of,  under  different  rates  of  change  of  angular  speed,  73;  in  a  circle, 
76,  154,  159;  of  a  point,  differential  equations  of,  81;  constrained,  of  a 
point,  88,  151-  rectilinear,  92;  simple  harmonic,  103;  in  resisting  medium, 
111;  of  projectiles,  117,  127;  compound  harmonic,  131;  planetary,  139,  141; 
on  an  inclined  plane,  151;  in  a  curved  path,  153;  in  a  cycloid,  155,  157; 
f  equations  of,  for  rotating  rigid  system,  176;  screw,  191;  components  of,  for 
a  rigid  system,  201;  relative,  of  a  body,  213;  relative  acceleration  of,  215. 


230  INDEX. 

Newton's  law  of  gravitation,  99. 

Normal,  acceleration,  52. 

Numeric,  of  a  quantity,  2;  equations  of  speed,  16;  equations  of  angular  speedy 

72;  equations  of  rate  of  change  of  speed,  25;  equations  of  rate  of  change  of 

angular  speed,  73. 

Oscillation,  amplitude  of,  104;  isochronous,  104. 

Paracentric  acceleration,  87. 

Parallel  angular  velocities,  centre  of,  192. 

Particle,  material,  14. 

Path,  of  a  point,  13;  curved,  motion  in,  153. 

Pendulum,  Blackburn's,  137;  simple,  154,  159. 

Per,  meaning  of,  3. 

Period,  in  harmonic  motion,  105. 

Phase,  in  harmonic  motion,  105. 

Physical  science,  1. 

Plane,  polar  co-ordinate,  10;  inclined,  motion  on,  151. 

Planetary  motion,  139;  path  of,  139;  application  of  calculus,  141. 

Point,  10;  position  of,  10;  of  reference,  10;  path  of,  13;  material,  14;  constrained 

motion  of,  88,  151;  kinematics  of,  91. 
Polar,  axis  of  earth,  length  of,  5,  9;  co-ordinates,  10;  equations  for  motion  of  a. 

point,  83. 
Polygon,  of  displacements,  35;  of  velocities,  43;  of  accelerations,  49. 
Position,  10;  of  a  point,  10. 
Poundal,  value  of,  9. 
Projectile,  motion  of,  117;  motion  of,  in  resisting  medium,  187. 

Quantity,  statement  of,  2;  vector,  34. 

Radian,  5;  square,  7;  solid,  7. 

Radius  vector,  11. 

Range  of  projectiles,  119,  127. 

Rate  of  change  of  speed,  24;  mean  and  instantaneous,  24;  a  scalar  quantity,  25; 
dimensions  of,  24;  unit  of,  24;  numeric  equations  of,  25;  sign  of,  25;  equa- 
tions of  motion  under  different  rates  of  change  of,  27;  graphic  rejiresenta- 
tion  of,  29. 

Rate  of  change  of  angular  speed,  73;  mean  and  instantaneous,  73;  unit  of,  72; 
numeric  equations  of,  73;  sign  of,  73;  equations  of  motion  under  different 
rates  of  change  of,  73;  in  terms  of  linear  speed,  75;  in  terms  of  moment 
of  tangential  acceleration,  76;  graphic  representation  of,  77. 

Rectangular  co-ordinates,  11. 

Rectilinear  motion,  92;  translation  and  rotation  combined,  193. 

Reference,  point  of,  10. 

Relative,  displacement,  34;  motion  of  a  body,  213;  acceleration  of  motion,  215. 

Relation  of  vertex  to  meridian,  5. 

Resistence,  coefficient  of,  in  resisting  medium.  111,  113. 

Resisting  medium,  motion  in,  111;  ceefficient  of  resistance  in.  111,  113;  motion 
of  projectiles  in,  127. 

Resolution  and  Composition,  of  displacements,  36;  of  velocities,  43;  of  acceler- 
ations, 49;  of  moments,  62;  of  angular  displacements,  171;  of  angular 
velocities  and  accelerations,  176;  of  translation  and  angular  displacement. 
187;  of  translation  and  angular  velocity  or  acceleration,  189;  of  screws,  203. 

Rest,  12;  absolute  and  relative,  12. 

Resultant,  angular  displacement,  velocity  or  acceleration,  178;  angular  velocity 
and  velocity  of  translation  for  a  rigid  system,  197. 

Rigid  system,  12;  kinematics  of,  169;  angular  displacement  of,  170;  line  repre- 
sentative of  angular  displacement  of,  170;  angular  velocity  of,  174;  equa- 
tions of  motion  of  rotating,  176;  displacement  of,  187;  resultant  angular 
velocity  and  velocity  of  translation  of,  197;  components  of  motion  of,  201; 
general  analytical  relations  of  rotating,  219. 

Rotation,  motion  of,  13,  169;  and  translation  combined,  13,  193;  axis  of,  173; 
instantaneous  axis  of,  174;  equations  of  motion  for,  176;  condition  for,  178J 
about  intersecting  axes,  195. 


VOL.   I. — KINEMATICS.  231 

■Science,  physical,  1. 

Screw  motion,  191;  composition  and  resolution  of,  203. 

Sign,  of  speed,  16;  of  rate  of  change  of  speed,  25;  of  components  of  displace- 
ment, 36;  of  components  of  velocity,  44,  of  components  of  acceleration,  50; 
of  components  of  moments,  62;  of  angular  speed,  72;  of  rate  of  change 
of  angular  speed,  73;  of  angular  displacement,  173;  of  angular  velocity, 
176;  of  angular  acceleration,  176;  of  moment  of  angular  velocity  or 
acceleration,  177. 

Simple  pendulum,  154;  application  of  calculus,  159. 

Solid  angle,  7;  radian,  7. 

Space,  dimensions  of,  12;  polar  co-ordinates,  10. 

Standard  units,  4;  yard,  4;  unit  of  length,  4;  unit  of  mass,  5;  unit  of  time,  4. 

States  of  matter.  1. 

Statement  of  a  quantity,  2. 

Statics,  definition  of,  1. 

Speed,  15;  mean  and  instantaneous,  15;  dimensions  of  unit  of,  15;  numeric  equa- 
tions of,  16;  sign  of,  16;  a  scalar  quantity,  16;  rate  of  change  of,  24;  mean 
and  instantaneous  rate  of  change  of,  24;  dimensions  of  unit  of  rate  of 
change  of,  24;  numeric  equations  of  rate  of  change  of,  25;  sign  of  rate  of 
change  of,  25;  angular,  72;  mean  and  instantaneous  angular,  72;  numeric 
equations  of  angular,  72;  sign  of  angular,  72;  rate  of  change  of  angular, 
73;  mean  and  instantaneous  rate  of  change  of  angular,  73;  numeric  equa- 
tions of  rate  of  change  of  angular,  73;  sign  of  rate  of  change  of  angular, 
73;  linear  in  terms  of  angular,  74;  angular  in  terms  of  moment  of  linear 
velocity,  75;  angular  rate  of  change  of,  in  terms  of  linear,  75;  angular  in 
terms  of  normal  acceleration,  76;  angular  rate  of  change  of,  in  terms  of 
moment  of  tangential  acceleration,  76;  graphic  representation  of  rate  of 
change  of  angular,  77. 

Square  radian,  7. 

System,  12;  rigid,  12;  material,  14;  kinematics  of  rigid,  169;  angular  displace- 
ment of  rigid,  170;  line  representative  of  angular  displacement  of  rigid, 
170;  angular  velocity  of  rigid,  174;  equations  of  motion  of  rotating  rigid, 
176;  displacement  of  rigid,  187;  resultant  angular  velocity  and  velocity  of 
translation  of  rigid,  197;  components  of  motion  of  rigid  201;  general 
analytical  relations  of  rigid  rotating,  219. 

Table  of  measures,  8. 

Tangential  acceleration,  52. 

Time,  unit  of,  4;  standard  unit  of,  4. 

Trajectory,  equation  of,  118;  velocity  at  any  point  of,  119;  time  of  flight,  119; 

horizontal  range,  119;  greatest  height,  120;  displacement  in  anv  direction, 

120;  angle  of  elevation,  120;  envelope  of  trajectories.  121. 
Translation,  motion  of,  13,  91;  and  rotation  combined,   13,  193:  curved  path, 

130;    and  angular  displacement,  187;  and  angular  velocity,  189;  resultant 

for  a  rigid  system,  197. 
Triangle  and  polygon,  of  displacements,  35;  of  velocities,  43;  of  accelerations,  49. 

Unit,  derived,  2;  of  measurement,  2;  of  length,  4,  8;  of  time,  4;  of  mass,  5,  9; 
of  angle,  5;  of  conical  angle,  6;  of  curvature,  8;  of  speed,  15;  of  rate  of 
change  of  speed,  24;  of  velocity,  43;  of  acceleration,  49;  of  angular  ve- 
locity, 174;  of  angular  acceleration,  176. 

Vector,  quantity,  34;  radius,  11.  j  i,i     ^o 

Velocity,  42;  mean  and  instantaneous,  42;  unit  of,  43;  uniform  and  variable,  46; 
line  representative  of,  43;  resolution  and  composition  of,  43;  rectangular 
components  of,  44;  sign  of  components  of,  44;  resultant,  45,  63;  moment 
of  60  177;  moment  of,  in  terms,  of  angular,  75;  couple,  81;  unit  of 
angular  174;  angular,  in  terms  of  linear,  176;  composition  and  resolution 
of  angular,  176;  concurring  angular,  178;  composition  and  resolution  of 
translation  and  angular,  189;  centre  of  parallel  angular,  192;  resultant 
angular  for  a  rigid  system,  197. 


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6 


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5  00 

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25 

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Bazin's  Experiments  upon  the  Contmction  of  the  Liquid  Vein. 

(Traulwine.) 8vo,  2  00 

Bovey's  Treatise  on  Hydraulics Svo,  4  00 

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Sbeep,  7  50 

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MANUFACTURES. 

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Allen's  Tables  for  Iron  Analysis Svo,  3  00' 

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Spencer's  Sugar  Manufacturer's  Handbook  . . .  .16mo,  morocco,  2  00 
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Thurston's  Manual  of  Steam  Boilers Svo,  5  00 

Walke's  Lectures  on  Explosives Svo,  4  00 

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Baker's  Masonry  Construction Svo,  5  00 

Beardslee  and  Kent's  Strength  of  Wrought  Iron Svo,  1  50 

10 


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15  00 

5  00 

6  OO 

10  00 

6  00 

7  50 

7  50 

5  00 

4  00 

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5  OO 

1  25 

2  00 

5  OO 

8  00 

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3  50 

2  50 

2  00 

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Byrne's  Highway  Construction 8vo, 

Church's  Mechanics  of  Engineering — Solids  and  Fluids 8vo, 

Du  Bois's  Stresses  in  Framed  Structures Small  4to, 

Johnson's  Materials  of  Construction 8vo, 

Lanza's  Applied  Mechanics 8vo, 

Martens's  Testing  Materials.     (Henning.) 2  vols.,  8vo, 

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Merriman's  Mechanics  of  Materials 8vo, 

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Rock-well's  Roads  and  Pavements  in  France 12mo, 

Spalding's  Roads  and  Pavements 12mo, 

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Wood's  Resistance  of  Materials 8vo, 

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Baker's  Elliptic  Functions 8vo,  1  50 

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Chapman's  Theory  of  Equations ISmO; 

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"        Primary  Geometry 12mo,  75 

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MECHANICS-MACHINERY. 

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12 


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3  50 

1  00 

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75 

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Lanza's  Applied  Mechanics gvo  7  50 

MacCord's  Kinematics 8vo  5  00 

Merriman's  Mechanics  of  Materials 8vo  4  00 

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3  50 

2  50 

1  50 

Chester's  Catalogue  of  Minerals 8vo, 

Paper, 

"       Dictionary  of  the  Names  of  Minerals 8vo, 

Dana's  American  Localities  of  Minerals, Large  8vo, 

"      Descriptive  Mineralogy.  (E.  S.)  Large  8 vo.  half  morocco, 
"      First  Appendix  to  System  of  Mineralogy.   . .  .Large  8vo, 

"      Mineralogy  and  Petrography.     (J.  D.) 12mo, 

"      Minerals  and  How  to  Study  Them.     (E.  S.) . .  12mo, 

"      Text-book  of  Mineralogy.    (E.  S.)... New  Edition.     8vo, 

*  Drinker's  Tunnelling,  Explosives,  Compounds,  and  Rock  Drills. 

4to,  half  morocco, 

Egleston's  Catalogue  of  Minerals  and  Synonyms 8vo, 

Eissler's  Explosives — Nitroglycerine  and  Dynamite 8vo, 

Hussuk's  Rock  forming  Minerals.     (Smith.) Small  8vo, 

Ihlseng's  Manual  of  Mining . . 8vo, 

Kuuhardt's  Ore  Dressing  in  Europe , 8vo, 

O'Driscoll's  Treatment  of  Gold  Ores 8vo, 

*  Penfield's  Record  of  Mineral  Tests Paper,  8vo, 

Roseubusch's    Microscopical    Physiography  of    Minerals    and 

Rocks.     (Iddings.) 8vo, 

Sawyer's  Accidents  in  Mines Ljirge  8vo, 

Stockbridge's  Rocks  and  Soils 8vo, 

Walke's  Lectures  on  Explosives 8vo, 

WilKams's  Lithology .>. jyt.-) '.)F» 8vo, 

Wilson's  Mine  Ventilation 12mo, 

"         Hj'draulic  and  Placer  Mining 12mo,       2  50 

STEAM  AND  ELECTRICAL  ENGINES,  BOILERS,  Etc. 

Stationaky — Makine — Locomotive — Gas  Engines,  Etc. 
{See  also  Engineeking,  p.  7.) 

Baldwin's  Steam  Heating  for  Buildings 12mo,  2  50 

Clerk's  Gas  Engine > Small  8vo,  4  00 

Ford's  Boiler  Making  for  Boiler  Makers 18mo,  1  00 

Heraen way 's  Indicator  Practice 12mo,  2  00 

Kneasss  Practice  and  Theory  of  the  Injector 8vo,  1  50 

MacCord's  Slide  Valve 8vo,  2  00 

Meyer's  Modern  Locomotive  Construction 4to,  10  00 

Peabody  and  Miller'3  Steam-boilers 8vo,  4  00 

Peabody's  Tables  of  Saturated  Steam 8vo,  1  00 

"         Thermodynamics  of  the  Steam  Engine 8vo,  5  00 

"         Valve  Gears  for  the  Steam  Engine 8vo,  2  50 

"           Steam-engine  Indicator 12mo,  1  50 

Pray's  Twenty  Years  with  the  Indicator Large  Svo,  2  50 

Pupin  and  Osterberg's  Tliermodynamics 12mo,  1  25 

14 


|1  25 

50 

3  00 

1  00 

12  50 

1  00 

2  00 

1  50 

4  00 

25  00 

2  50 

4  00 

2  00 

4  00 

1  50 

2  00 

50 

5  00 

7  00 

2  50 

4  00 

3  00 

1  25 

6  00 

10  00 

75 

1  50 

2  50 

5  00 

2  50 

5  00 

5  00 

2  50 

4  00 

Reagan's  Steam  and  Electric  Locomotives 12mo,     |2  00 

Eoiitgeu's  Theimodyuamics.     (Du  Bois. ) 8vo        5  00 

Sinclair's  Locomotive  Running , 12mo       2  00 

Snow's  Steam-boiler  Practice 8vo.      3  00 

Thurston's  Boiler  Explosions 12mo,      1  50 

"  Engine  and  Boiler  Trials 8vo,       5  00 

"  Manual  of  the  Steam  Engine.      Part  I.,  Structure 

and  Theory 8vo,      6  00 

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Construction,  and  Operation 8vo, 

2  parts, 

Thurston's  Philosophy  of  the  Steam  Engine 12mo, 

"  Reflection  on  the  Motive  Power  of  Heat.    (Caruot.) 

12mo. 

"  Stationarj'-  Steam  Engines 8vo, 

"  Steam-boiler  Construction  and  Operation 8vo, 

Spaugler's  Valve  Gears 8vo, 

Weisbach's  Steam  Engine.     (Du  Bois.)li  .iWA  ./f.W.. .  .8vo, 

Whitham's  Steam-engine  Design 8vo, 

"Wilson's  Steam  Boilers.     (Flather.) 12rao, 

Wood's  Thermodynamics,  Heat  Motors,  etc 8vo, 

TABLES,  WEIGHTS,  AND  MEASURES. 

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Crandnll's  Railway  and  Earth vrork  Tables 8vo,  1  50 

Egleston's  Weights  and  Measures 18mo,  75 

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15 


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Ferrel's  Treatise  ou  tlie  Winds 8vo,  4  00 

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Rothcrham's    The    New    Testament     Criticall}'    Emphasized. 

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"       Hebrew  Chrestomathy ; 8vo,  2  00 

Letteris's    Hebrew   Bible  (Massoretic   Notes  in  English). 

'                                                                                8vo,  arabesque,  2  25 

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Hammarsten's  Physiological  Chemistry.   (Mandel.) 8vo,  4  00 

Mott's  Composition,  Digestibility,  and  Nutritive  Value  of  Food. 

Large  mounted  chart,  1  25 

Ruddiman's  Incompatibilities  in  Prescriptions 8vo,  2  00 

Steel's  Treatise  on  the  Diseases  of  the  Ox 8vo,  6  00 

"      Treatise  on  the  Diseases  of  the  Dog 8vo,  3  50 

Woodhull's  Military  Hygiene 16mo,  1  50 

Worcester's  Small  Hospitals — Establishment  and  Maintenance, 
including  Atkinson's  Suggestions  for  Hospital  Archi- 
tecture   .12mo,  I  25 

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